Under choice, a space is metrizable if and only if it is regular, $T_1$, and has a $\sigma$-locally-finite basis

If $X$ is metrizable, it is regular and $T_1$, and [L1] gives the required basis.

Conversely write the supplied basis as $\mathcal{B}={\bigcup }_{n\ge 0}{\mathcal{B}}_n$ with every ${\mathcal{B}}_n$ locally finite. We first prove normality, rather than assuming it. For disjoint closed $F,H$, let $U_n$ be the union of those $B\in {\mathcal{B}}_n$ with $\bar{B}\cap H=\mathrm{\varnothing }$, and define $V_n$ symmetrically for $H$ against $F$. Regularity and the basis property say the $U_n$ cover $F$ and the $V_n$ cover $H$: each point outside a closed set has a basis neighborhood whose closure misses it. By [L2], $\overline{U_n}$ misses $H$ and $\overline{V_n}$ misses $F$. Now put $U={\bigcup }_{n\ge 0}(U_n\setminus {\bigcup }_{k\le n}\overline{V_k}),V={\bigcup }_{n\ge 0}(V_n\setminus {\bigcup }_{k\le n}\overline{U_k})\mathrm{.}$ Both are open, $F\subseteq U$, and $H\subseteq V$. If a point belonged to the $n$th piece of $U$ and the $m$th piece of $V$, then $m\le n$ would contradict its exclusion from $\overline{V_m}$, while $n\le m$ would contradict its exclusion from $\overline{U_n}$. Thus $U\cap V=\mathrm{\varnothing }$ and $X$ is normal.

Every basis member $B$ is the cozero set of a continuous function. For each $n$, let $F_{B,n}$ be the union of $\bar{C}$ over $C\in {\mathcal{B}}_n$ with $\bar{C}\subseteq B$. It is closed and contained in $B$ by [L2]. The union of these $F_{B,n}$ over $n$ is $B$: for $x\in B$, regularity gives an open $O$ with $x\in O\subseteq \bar{O}\subseteq B$, and some basis member $C\ni x$ lies in $O$. By [L3] applied to the disjoint closed sets $F_{B,n}$ and $X\setminus B$, choose $f_{B,n}:X\to [0,1]$ equal to $1$ on $F_{B,n}$ and $0$ outside $B$. AC makes these simultaneous choices for all $(B,n)$. The uniformly convergent sum $u_B={\sum }_{n\ge 0}{2}^{-n-1}{f}_{B,n}$ is continuous, zero outside $B$, and positive at every point of $B$. Hence $B=\{u_B>0\}$.

Index repeated basis members by their layer, writing $i=(n,B)$ with $B\in {\mathcal{B}}_n$. At each $x$, only finitely many members of ${\mathcal{B}}_n$ contain $x$, so the finite sum $S_n(x)={\sum }_{B\in {\mathcal{B}}_n}{u}_B(x{)}^2$ is well defined. It is continuous: around any $x$, local finiteness makes all but finitely many summands identically zero. Define $a_{nB}(x)=2^{-n-1}\frac{u_B(x)}{\sqrt{1+S_n(x)}}\mathrm{.}$ Each coordinate is continuous and ${\sum }_{B\in {\mathcal{B}}_n}{a}_{nB}(x{)}^2\le 4^{-n-1}$ uniformly in $x$. Consequently the nonnegative generalized sum $d(x,y{)}^2={\sum }_{n\ge 0}{\sum }_{B\in {\mathcal{B}}_n}(a_{nB}(x)-a_{nB}(y){)}^2$ is finite. It is the squared ${\mathrm{\ell }}^2$ distance of the two coordinate vectors; the triangle inequality follows from finite-sum Cauchy--Schwarz and passage to the supremum over finite index sets. Symmetry and $d(x,x)=0$ are immediate. If $x\mathrm{\ne }y$, $T_1$ and the basis give a $B$ with $x\in B$, $y\mathrm{\notin }B$; then $a_{nB}(x)>0=a_{nB}(y)$ for a layer containing $B$. Thus $d$ is a metric.

The metric induces the original topology. Given $x\in O$ open, choose a basis member $B$ with $x\in B\subseteq O$ and a layer $n$ containing it. If $d(x,y)<a_{nB}(x)\mathrm{/}2$, then $a_{nB}(y)>0$, so $y\in B\subseteq O$. Conversely fix $x$ and $\epsilon >0$. Choose $N$ so that $4{\sum }_{n>N}{4}^{-n-1}<{\epsilon }^2\mathrm{/}2$. For each of the finitely many layers $n\le N$, take a neighborhood of $x$ meeting only finitely many $B\in {\mathcal{B}}_n$. On the intersection of these neighborhoods, every other coordinate in those layers vanishes both at $x$ and at nearby $y$. Continuity of the remaining finitely many coordinates makes their squared-difference sum $<{\epsilon }^2\mathrm{/}2$ on a smaller neighborhood. The tail estimate $(a-b{)}^2\le 2a^2+2b^2$ and the uniform layer bound make its contribution $<{\epsilon }^2\mathrm{/}2$. Thus $d(x,y)<\epsilon$ there.

Step 1.1 proves the forward implication, and steps 1.2--3.1 construct a compatible metric for the reverse implication under the stated AC.