Under choice, metric spaces have sigma-discrete open bases

Fix the well-order in A1. For $k,n\in \mathbb{N}$ and $a\in X$, put $U_{a,k}=B_d(a,2^{-k})$, $r_n=2^{-n}$, and define $F_{a,k,n}=\{z\in X:(\mathrm{\forall }y\in X\setminus U_{a,k})d(z,y)\ge r_n\}\setminus {\bigcup }_{b<a}{U}_{b,k}\mathrm{.}$ The first set is closed, being the intersection over $y\mathrm{\notin }{U}_{a,k}$ of the closed sets $\{z:d(z,y)\ge r_n\}$; the triangle inequality makes their complements open. Its defining condition is vacuous if $U_{a,k}=X$. The subtracted union is open, so $F_{a,k,n}$ is closed, and it lies in $U_{a,k}$ since a point outside that set violates the condition with $y=z$.

For fixed $k,n$ the nonempty cores are pairwise $r_n$ separated. Indeed, if $a<b$, $u\in F_{a,k,n}$ and $v\in F_{b,k,n}$, then $v\mathrm{\notin }{U}_{a,k}$ by the subtraction defining the latter core, and hence $d(u,v)\ge r_n$. For each fixed $k$ their union over $a,n$ is $X$: given $x$, the set of centers $a$ with $x\in U_{a,k}$ is nonempty (it contains $x$), so has a least member $a$. Openness gives some $\delta >0$ with $B_d(x,\delta )\subseteq U_{a,k}$. Take $n$ with $r_n\le \delta$. Then every $y\mathrm{\notin }{U}_{a,k}$ satisfies $d(x,y)\ge r_n$, and $x$ lies in none of the earlier balls, so $x\in F_{a,k,n}$. These are least selections or existential instantiations, requiring no further choice.

For each nonempty core define the open set $V_{a,k,n}={\bigcup }_{u\in F_{a,k,n}}{B}_d(u,r_n\mathrm{/}3)$. It contains its core and is contained in $U_{a,k}$: a point outside $U_{a,k}$ has distance at least $r_n$ from every core point. For fixed $k,n$, every ball $B_d(x,r_n\mathrm{/}6)$ meets at most one of these sets. Otherwise two points of that ball belonging to different $V$ sets yield corresponding core points $u,v$ with $d(u,x)<r_n\mathrm{/}2$ and $d(v,x)<r_n\mathrm{/}2$, whence $d(u,v)<r_n$, contradicting step 2.1. Thus each layer ${\mathcal{V}}_{k,n}=\{V_{a,k,n}:F_{a,k,n}\mathrm{\ne }\mathrm{\varnothing }\}$ is discrete, and its union over $n$ covers $X$ for every $k$.

The union of all layers is a basis. If $x\in O$ with $O$ open, choose $\epsilon >0$ with $B_d(x,\epsilon )\subseteq O$ and $k$ with $2^{1-k}<\epsilon$. By step 3.1 some $V_{a,k,n}$ contains $x$. Both $x$ and each $z$ in that set lie in $U_{a,k}$, so $d(x,z)<2^{1-k}<\epsilon$; therefore $x\in V_{a,k,n}\subseteq O$. Enumerate pairs $(k,n)$ by successive finite diagonals $k+n=0,1,2,\dots$ to obtain a sequence of discrete layers. If $X$ is empty, all layers are empty and the same basis criterion holds vacuously. This proves the claimed $\sigma$-discrete open basis, with AC used only for the initial well-order.