Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line

For $x,y\in {\mathbb{R}}^n$ and $k<n$ the term $(x_k-y_k{)}^2$ is one of the nonnegative terms of ${\sum }_{j<n}(x_j-y_j{)}^2$, so $(x_k-y_k{)}^2\le d_2(x,y{)}^2$, and taking nonnegative square roots gives $\mathrm{\mid }{x}_k-y_k\mathrm{\mid }\le d_2(x,y)$; hence $d_{\mathrm{\infty }}(x,y)\le d_2(x,y)$.

Conversely each $(x_j-y_j{)}^2\le d_{\mathrm{\infty }}(x,y{)}^2$, so $d_2(x,y{)}^2\le \iota (n)d_{\mathrm{\infty }}(x,y{)}^2\le (\iota (n)d_{\mathrm{\infty }}(x,y){)}^2$, the last step because $\iota (n)\ge 1$; hence $d_2(x,y)\le \iota (n)d_{\mathrm{\infty }}(x,y)$.

For claim 1 fix reals $a_k\le b_k$ $(k<n)$ and the box $Q$ they determine, let $(U_i{)}_{i\in I}$ be open subsets of ${\mathbb{R}}^n$ with $Q\subseteq {\bigcup }_{i\in I}{U}_i$, call a set $S\subseteq {\mathbb{R}}^n$ finitely covered when finitely many of the $U_i$ have union containing $S$, and suppose for contradiction that $Q$ is not finitely covered.

For a box $P=\{x:c_j\le x_j\le e_j(j<n)\}$ with $c_j\le e_j$ and for $k<n$, let $P^{k,0}$ and $P^{k,1}$ be the boxes obtained by replacing the $k$-th interval $[c_k,e_k]$ by $[c_k,(c_k+e_k)\mathrm{/}2]$ and by $[(c_k+e_k)\mathrm{/}2,e_k]$; then $P=P^{k,0}\cup P^{k,1}$ by trichotomy applied to $x_k$ against the midpoint, the $k$-th side length of each is $(e_k-c_k)\mathrm{/}2$ and the others are unchanged, and if both halves were finitely covered so would $P$ be, the union of two finite subfamilies being finite. Define $H_k(P):=P^{k,0}$ if $P^{k,0}$ is not finitely covered, and $H_k(P):=P^{k,1}$ otherwise; this is a definition by a property, and $H_k(P)$ is not finitely covered whenever $P$ is.

Recursion on $\mathbb{N}\times Z$, with $Z$ the set of functions from boxes to boxes, starting value $(0,\mathrm{i}\mathrm{d})$ and rule $(j,h)\mapsto (j+1,H_j\circ h)$ for $j<n$ and $(j,h)\mapsto (j+1,h)$ otherwise, produces $G_j$ for every $j$; put $G:=G_n$. By induction on $j\le n$, $G_j(P)\subseteq P$ is a box whose $k$-th side is half that of $P$ for $k<j$ and equal to that of $P$ for $k\ge j$, and $G_j(P)$ is not finitely covered when $P$ is not. So $G(P)\subseteq P$ halves every side and preserves not being finitely covered.

Recursion applied to the starting value $Q$ and the rule $G$ produces boxes $P_m$ with $P_0=Q$ and $P_{m+1}=G(P_m)$; each $P_m$ fails to be finitely covered, $P_{m+1}\subseteq P_m$, and the $k$-th side length of $P_m$ is ${\mathrm{\ell }}_k(1\mathrm{/}2{)}^m$, where ${\mathrm{\ell }}_k:=b_k-a_k\ge 0$.

For each $k<n$ the $k$-th intervals of the $P_m$ form a nested family of closed bounded intervals whose lengths ${\mathrm{\ell }}_k(1\mathrm{/}2{)}^m$ tend to $0$, so their intersection is a single point $p_k$; the function $p:n\to \mathbb{R}$, $k\mapsto p_k$, is a point of ${\mathbb{R}}^n$ lying in every $P_m$.

Since $p\in P_0=Q\subseteq {\bigcup }_{i\in I}{U}_i$, there is $i^{\ast }\in I$ with $p\in U_{i^{\ast }}$, and openness gives a real $r>0$ with $B(p,r)\subseteq U_{i^{\ast }}$.

Put $L:=\max \{{\mathrm{\ell }}_k:k<n\}\ge 0$ and $C:=\iota (n)L+1>0$; for $x\in P_m$ each $\mathrm{\mid }{x}_k-p_k\mathrm{\mid }$ is at most the $k$-th side length of $P_m$, so $d_{\mathrm{\infty }}(x,p)\le L(1\mathrm{/}2{)}^m$ and $d_2(x,p)\le \iota (n)L(1\mathrm{/}2{)}^m\le C(1\mathrm{/}2{)}^m$ by step 1.2. Taking a natural $N\ge 1$ with $1\mathrm{/}N<r\mathrm{/}C$ and then $m$ with $(1\mathrm{/}2{)}^m<1\mathrm{/}N$ gives $P_m\subseteq B(p,r)\subseteq U_{i^{\ast }}$, so $P_m$ is finitely covered by the single set $U_{i^{\ast }}$, contradicting step 5.1.

Therefore every such family has finitely many members covering $Q$, and $Q$ is a compact subset of $({\mathbb{R}}^n,d_2)$: claim 1 is proved.

For claim 2, a compact $K\subseteq {\mathbb{R}}^n$ is closed and bounded.

Conversely let $K\subseteq {\mathbb{R}}^n$ be closed and bounded; if $K=\mathrm{\varnothing }$ it is compact, and otherwise $K\subseteq B(x_0,\rho )$ for some $x_0$ and real $\rho >0$, so every $x\in K$ and $k<n$ satisfy $\mathrm{\mid }{x}_k\mathrm{\mid }\le \mathrm{\mid }(x_0{)}_k\mathrm{\mid }+\mathrm{\mid }{x}_k-(x_0{)}_k\mathrm{\mid }\le \mathrm{\mid }(x_0{)}_k\mathrm{\mid }+d_2(x,x_0)<\mathrm{\mid }(x_0{)}_k\mathrm{\mid }+\rho$ by step 1.1; with $M:=\max \{\mathrm{\mid }(x_0{)}_k\mathrm{\mid }:k<n\}+\rho$ the box $Q_M:=\{x:-M\le x_k\le M(k<n)\}$ contains $K$.

$K$ is the trace on $Q_M$ of a closed subset of ${\mathbb{R}}^n$, namely of $K$ itself, so $K$ is closed in the metric subspace $Q_M$; that subspace is compact by step 9.1, so $K$ is compact, and claim 2 is proved.

For claim 3, let $\psi :\mathbb{R}\to {\mathbb{R}}^1$ send $t$ to the function $1\to \mathbb{R}$ with value $t$; it is a bijection and $d_2(\psi (s),\psi (t))=\sqrt{(s-t{)}^2}=\mathrm{\mid }s-t\mathrm{\mid }=d_{\mathbb{R}}(s,t)$, so $\psi$ carries each ball onto the corresponding ball, hence open sets onto open sets and open covers onto open covers with matching finite subfamilies, and likewise closed sets onto closed sets and bounded sets onto bounded sets. Applying claim 2 with $n=1$ to $\psi [K]$ therefore gives claim 3.