Bing's Q-set space is a normal nonmetrizable Moore space

Fix $E$, $Z$, $P$, $E_0$ and the covers ${\mathcal{H}}_n$.

Each ${\mathcal{H}}_n$ is an open cover by open sets, because $z\in V(z,n)$ and each $V(z,n)$ is a basic open set by [F1].

The zero-based sequence $({\mathcal{G}}_m{)}_{m\in \mathbb{N}}$ is a development, with positive scale $n=m+1$. At $z=(a,0)\in E_0$, if $q\in Z$ satisfies $z\in V(q,n)$ and $q\mathrm{\ne }z$, then $q\in P$ and $\mathrm{dist}(q,z)\ge q_2$, while $z\in V(q,n)$ forces $\mathrm{dist}(q,z)<q_2\mathrm{/}3$, a contradiction; hence $\mathrm{St}(z,{\mathcal{H}}_n)=U(z,n)$, and the tangent disks form a local base at $z$ by [F1], so every neighbourhood of $z$ contains a star. At $z\in P$, every member $V(q,n)$ containing $z$ satisfies $V(q,n)\subseteq B(z,2\mathrm{/}n)$ when $q\in P$, and $V(q,n)\subseteq B(z,4\mathrm{/}n)$ when $q\in E_0$; hence $\mathrm{St}(z,{\mathcal{H}}_n)\subseteq B(z,4\mathrm{/}n)$, which is contained in any prescribed Euclidean neighbourhood for $n$ large.

$Z$ is separable: the set of points of $P$ with both coordinates rational is countable and dense: every ordinary ball in $P$ contains a rational point, and the open disk part of $U((a,0),n)$ contains $(a,1\mathrm{/}(2n))$, hence also a rational point of $P$ sufficiently close to it (Separability: the existence of an at most countable dense subset, Open ball, closed ball and sphere in a metric space).

$E_0$ is closed in $Z$ and discrete: it is the trace of the closed set $\mathbb{R}\times \{0\}$, and each of its points $z$ has the neighbourhood $U(z,1)$ with $U(z,1)\cap E_0=\{z\}$ by [F1].

(Axis separation.) Identify the axis with $E$ via $a\mapsto (a,0)$ for this step. Let $A\subseteq E$ and put $B:=E\setminus A$; symbols for their neighbourhoods refer to the corresponding axis points. Since every subset of the Q-set $E$ is relatively $G_{\delta }$ there are decreasing sequences $(S_n)$ and $(T_n)$ of open subsets of $\mathbb{R}$ with $A=E\cap {\bigcap }_n{S}_n$ and $B=E\cap {\bigcap }_n{T}_n$, so $A\subseteq S_n$ and $B\subseteq T_n$ for every $n$; for $n\in \mathbb{N}$ put $B_n:=B\setminus S_n$ and $A_n:=A\setminus T_n$. Fix $n$. For $a\in A$ the set $S_n$ is open and contains $a$, so let $l_n(a)$ be the least positive integer with $(a-1\mathrm{/}{l}_n(a),a+1\mathrm{/}{l}_n(a))\subseteq S_n$, and put ${\epsilon }_n(a)=1\mathrm{/}{l}_n(a)$. This is a positive real even when $S_n=\mathbb{R}$. Every $b\in B_n=E\setminus S_n$ has $\mathrm{\mid }a-b\mathrm{\mid }\ge {\epsilon }_n(a)$. Let $k_n(a)$ be the least positive integer with $2\mathrm{/}\sqrt{k_n(a)}\le {\epsilon }_n(a)$. Then $U((a,0),k_n(a))\cap U((b,0),1)=\mathrm{\varnothing }$ for every $a\in A$ and $b\in B_n$: tangent disks of radii $1\mathrm{/}k$ and $1$ at axis points at distance $\rho$ are disjoint whenever $\rho \ge 2\mathrm{/}\sqrt{k}$, because their centre distance $\sqrt{{\rho }^2+(1-1\mathrm{/}k{)}^2}$ is then at least $1+1\mathrm{/}k$. Hence the open sets $V_n:={\bigcup }_{b\in B_n}U((b,0),1)$ and $W_n:={\bigcup }_{a\in A_n}U((a,0),k_n(a))$ satisfy $B_n\subseteq V_n$, $A_n\subseteq W_n$, $U((a,0),k_n(a))\cap V_n=\mathrm{\varnothing }$ for $a\in A$, $V_n\cap E_0=B_n$, $W_n\cap E_0=A_n$, and $V_n\cap W_n=\mathrm{\varnothing }$: a tangent disk meets the axis only at its own tangency point, and every disk occurring in $W_n$ was chosen disjoint from $V_n$. Put $V:={\bigcup }_n(V_n\setminus {\bigcup }_{k<n}\mathrm{cl}(W_k))$ and $W:={\bigcup }_n(W_n\setminus {\bigcup }_{k<n}\mathrm{cl}(V_k))$. Both are open and they are disjoint. If $x\in V_n\setminus {\bigcup }_{k<n}\mathrm{cl}(W_k)$ and $x\in W_m\setminus {\bigcup }_{k<m}\mathrm{cl}(V_k)$ with $n<m$, then $x\in V_n\subseteq {\bigcup }_{k<m}\mathrm{cl}(V_k)$, contradicting the choice of $x\in W_m$; symmetrically for $m<n$; and $n=m$ is excluded by $V_n\cap W_n=\mathrm{\varnothing }$. Also $B\subseteq V$ and $A\subseteq W$: for $b\in B$ the fact that $b\mathrm{\notin }A=E\cap {\bigcap }_n{S}_n$ gives an $m$ with $b\mathrm{\notin }{S}_m$, that is $b\in B_m\subseteq V_m$, while $b\mathrm{\notin }{\bigcup }_{k<m}\mathrm{cl}(W_k)$ because $\mathrm{cl}(W_k)\cap E_0=A_k\subseteq A$ is disjoint from $B$; symmetrically for $a\in A$. The closure identities $\mathrm{cl}(W_k)\cap E_0=A_k$ and $\mathrm{cl}(V_k)\cap E_0=B_k$ hold because $A_k=E\setminus T_k$ and $B_k=E\setminus S_k$ are closed in the Euclidean subspace $E$, not merely in the discrete axis: if a sequence of points of $W_k$ converges to $(c,0)\in E_0$, then its tangency points converge to $c$, since a point $(x,y)\in U((a,0),r)$ with $r\le 1$ satisfies $\mathrm{\mid }x-a{\mathrm{\mid }}^2<2ry-y^2$, so $\mathrm{\mid }x-a\mathrm{\mid }\to 0$ as $y\to 0$; hence $c$ lies in the Euclidean relatively closed set $A_k$, and dually for $B_k$.

Reduction to arbitrary closed sets. Let $C,D$ be disjoint closed subsets of $Z$. By step 2.5 choose disjoint open $O,Q$ containing $C\cap E_0$ and $E_0\setminus C$, respectively. For each $a\in C\cap E_0$ choose the least positive $n(a)$ with $U(a,n(a))\subseteq O\setminus D$, and put $R_C={\bigcup }_{a\in C\cap E_0}U(a,2n(a))$. Its closure misses $D\cap E_0$, since $R_C\subseteq O$ and $D\cap E_0\subseteq Q$. Its closure also misses $D\cap P$. Indeed, if a point $q=(x,y)$ lies in the disk of radius $r\mathrm{/}2$, tangent at $a$, where $r=1\mathrm{/}n(a)\le 1$, then $(x-a_1{)}^2+y^2<ry$, so its distance to the complement of the disk of radius $r$ is at least $y\mathrm{/}2$: the distance to the outer centre has square less than $r^2-ry$, and $r-\sqrt{r^2-ry}=ry\mathrm{/}(r+\sqrt{r^2-ry})\ge y\mathrm{/}2$. The outer disk misses $D$. A sequence in $R_C$ converging to $d\in D\cap P$ would eventually have height at least $d_2\mathrm{/}2$, and therefore distance at least $d_2\mathrm{/}4$ from $D$, impossible. A closure point in $P$ supplies such a sequence by its ordinary ball base (AC is available). Thus $\overline{R_C}\cap D=\mathrm{\varnothing }$. Reversing the roles, using $Q$ around $D\cap E_0$ and $O$ around $C\cap E_0$, gives open $R_D$ with $D\cap E_0\subseteq R_D$ and $\overline{R_D}\cap C=\mathrm{\varnothing }$.

$Z$ is $T_1$ and regular: distinct points are separated by small balls or tangent disks, using that $E_0$ is closed and $P$ is open; for $z=(a,0)\in E_0$ and a closed $C\ni ̸z$, choose $n$ with $U(z,n)\cap C=\mathrm{\varnothing }$ by [F1], let $D$ be the closed disk of radius $1\mathrm{/}n$ about $(a,1\mathrm{/}n)$ and put $U:=U(z,2n)$ and $V:=Z\setminus \mathrm{cl}(U)$; every point of $\mathrm{cl}(U)$ other than $z$ lies in $U(z,n)\subseteq Z\setminus C$ and $z\mathrm{\notin }C$, so $C\subseteq V$, while $U\cap V=\mathrm{\varnothing }$. Together with step 2.2 the space $Z$ is a Moore space.

$Z$ is not metrizable. Suppose $d$ induces its topology. By step 2.3 the metric space is separable, so by [L2] it has a countable base, say $(W_i{)}_{i\in \omega }$. For $z\in E_0$ the set $U(z,1)$ is open and contains $z$, so by [F1] some $i$ has $z\in W_i\subseteq U(z,1)$, and then $W_i\cap E_0=\{z\}$ by step 2.4; the map $z\mapsto \min \{i:z\in W_i\subseteq U(z,1)\}$ is therefore a definable injection $E_0\to \omega$, contradicting the uncountability of $E$ and hence of $E_0$.

Every point of $C\cap P$ lies in an ordinary ball with rational centre and rational positive radius whose Euclidean closed ball is contained in $P\setminus D$: first take a sufficiently small ball inside that open set, then a rational centre sufficiently close to the point and a rational radius between the required bounds. Such a closed ball has positive distance from the axis and is closed in $Z$. The family of all such rational balls is at most countable and covers $C\cap P$; list it with empty sets as padding, and prepend $R_C$. This gives $(I_n{)}_{n\in \mathbb{N}}$ covering $C$ with $\overline{I_n}\cap D=\mathrm{\varnothing }$. Similarly obtain $(J_n)$ covering $D$ with $\overline{J_n}\cap C=\mathrm{\varnothing }$, starting with $R_D$. The sets $I={\bigcup }_n(I_n\setminus {\bigcup }_{k\le n}\overline{J_k})$ and $J={\bigcup }_n(J_n\setminus {\bigcup }_{k<n}\overline{I_k})$ are open, contain $C,D$, and are disjoint: for a point in the terms indexed by $n,m$, if $m\le n$ the first term excludes $J_m$, and if $n<m$ the second excludes $I_n$. This includes empty traces and proves normality by [F4].

Steps 2.2, 3.2, 2.3, 3.3 and 4.1 show that $Z$ is a Moore space, separable, nonmetrizable and normal, as claimed.