A compact metric space has a finite Borel nearest-center quantization at every positive scale

For any $x\in X$, the ball $B(x,\epsilon )$ is open: if $y\in B(x,\epsilon )$, set $\delta =\epsilon -d(x,y)>0$; whenever $d(y,z)<\delta$, [F2] gives $d(x,z)\le d(x,y)+d(y,z)<\epsilon$, so $B(y,\delta )\subseteq B(x,\epsilon )$. Also $x\in B(x,\epsilon )$, so $\{B(x,\epsilon ):x\in X\}$ is an open cover.

By [F1], finitely many members of that cover still cover $X$. Each selected member is a ball with at least one center; since the list is finite, finite induction selects a center for every selected member without any choice axiom. Removing repeated centers and preserving their first-occurrence order leaves a finite nonempty distinct list $x_1,\dots ,x_m$ whose $\epsilon$-balls cover $X$.

For $i<j$, the function $h_{ij}(x)=d(x,x_i)-d(x,x_j)$ satisfies $\mathrm{\mid }{h}_{ij}(x)-h_{ij}(y)\mathrm{\mid }\le 2d(x,y)$ by applying the triangle inequality in both directions to each distance. If $h_{ij}(x)<0$, then the ball of radius $-h_{ij}(x)\mathrm{/}4$ around $x$ lies in $\{h_{ij}<0\}$; if $h_{ij}(x)>0$, then the ball of radius $h_{ij}(x)\mathrm{/}4$ around $x$ lies in $\{h_{ij}>0\}$. Hence $\{h_{ij}<0\}$ is open and $\{h_{ij}\le 0\}$ is closed.

Define $C_i:={\bigcap }_{j<i}\{x:d(x,x_i)<d(x,x_j)\}\cap {\bigcap }_{j>i}\{x:d(x,x_i)\le d(x,x_j)\}$. These are Borel sets by [F4] and step 3.1. For each $x$, [F3] gives a least index minimizing the finite list $d(x,x_1),\dots ,d(x,x_m)$, so $x$ belongs to the corresponding $C_i$. If $i<j$, membership in $C_i$ gives $d(x,x_i)\le d(x,x_j)$ while membership in $C_j$ gives $d(x,x_j)<d(x,x_i)$; therefore the cells are disjoint and form a partition.

Define $Q(x)=x_i$ on $C_i$. For every subset $A\subseteq F$, $Q^{-1}(A)={\bigcup }_{i:x_i\in A}{C}_i$ is a finite union of Borel sets, so $Q$ is Borel. The net points from step 2.1 have $\epsilon$-balls covering $X$; hence a nearest center is at distance less than $\epsilon$ from every point. Thus $d(x,Q(x))<\epsilon$ for all $x$.