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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?

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    $\begingroup$ I don't get your question. What is the statement you want to prove without choice? $\endgroup$ – Asaf Karagila Oct 6 '18 at 17:12
  • $\begingroup$ Passing from a generic open cover to a ball cover. Let me rephrase it in a clearer way $\endgroup$ – Manlio Oct 7 '18 at 9:59
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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 $A$.

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