Under choice, every open cover of a metric space has a point-finite open refinement

Well order $\{C_\alpha\}$ by [A1], and write $R(x,n)=B(x,2^{-n})$. A ball $R(z,n+1)$ is chosen for $C_\alpha$ when $n$ is the least natural number with $R(z,n)\subseteq C_\alpha$ and, in addition, $R(z,n)\subseteq C_\beta$ for some $\beta<\alpha$. Let $\mathcal G_\alpha$ be the union of all balls chosen for $C_\alpha$.

Put $C'_\alpha:=C_\alpha\setminus\overline{\mathcal G_\alpha}$. Each $C'_\alpha$ is open and refines $C_\alpha$.

The $C'_\alpha$ cover. Otherwise let $C_\alpha$ be the first original member containing an omitted point $x$. Then $x\in\overline{\mathcal G_\alpha}$. By [L1], choose $N$ with $B(x,3\cdot2^{-N})\subseteq C_\alpha$, and put $\delta=2^{-(N+2)}$. Some chosen ball $R(z,n_z+1)$ meets $B(x,\delta)$; write its radius as $r=2^{-(n_z+1)}$. If $r>\delta$, then $d(x,z)<r+\delta<2r$, so its expanded ball $R(z,n_z)$ contains $x$. If $r\le\delta$, then $d(x,z)<r+\delta\le2\delta<2^{-N}$, so $R(z,N)\subseteq C_\alpha$ and minimality gives $n_z\le N$; hence $r\ge2^{-(N+1)}=2\delta$, a contradiction. Thus in every case an expanded chosen ball contains $x$. That expanded ball lies in some $C_\beta$ with $\beta<\alpha$, contradicting the choice of $\alpha$.

If $x\in C'_\alpha$ and $n$ is least with $R(x,n)\subseteq C_\alpha$ (which exists by [L1]), then $C_\alpha$ is the first cover member containing $R(x,n)$: otherwise $R(x,n+1)$ would be chosen for $C_\alpha$ and would contain $x$, contrary to $x\notin\overline{\mathcal G_\alpha}$. For each $n$ there is at most one such first member, and as $n$ increases their ordinal indices are nonincreasing. Infinitely many distinct indices would therefore give an infinite strictly descending sequence of ordinals, impossible because its range has a least member. Thus only finitely many $C'_\alpha$ contain $x$.

Thus $\{C'_\alpha\}$ is the point-finite open refinement required by [F2].