Let $X$ be a metric space and assume that, for every $\varepsilon>0$ there is a countable open cover $(A_i)$ of $X$ with $diam(A_i)\le \varepsilon$ for each $i$.
Of course I can assume the cover is made only of basic open sets: if $\varepsilon$ is fixed then, for each $i$, we can pick $x_i\in A_i$ and consider the ball $B_i$ with center $x_i$ and diameter $2\varepsilon$ to obtain a $2\varepsilon$-cover of $X$.
The question is: can I assume the open cover is made of open balls even without $AC$ (or, in this case, without $AC(\omega)$)? Is there a way to build a ball cover from a generic open cover that is not dependent on the choice of a point for each set in the cover?
The answer is unfortunately negative, at least if you want to keep the cover countable.
In Cohen's first model there is an infinite Dedekind-finite set of reals which is dense in $\Bbb R$. This is a set which is infinite, but has no countably infinite subset. And as a set of reals, it is of course a metric space, and by being dense, it is also not bounded.
Now, cover $A$ with rational intervals (or their intersection with $A$) with diameter $\varepsilon$, this is a countable cover. But if you want these to be open balls of diameter $\varepsilon$, you need to choose centers for these balls, which would necessitate a countably infinite subset of $A$.