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Let $X$ be a metric space. I want to show that:

If a subset $A \subset X$ is totally bounded, then its closure $\overline{A}$ is totally bounded.

Definition of "totally bounded": A set $A$ is totally bounded if, for each $\varepsilon > 0$, there is a finite $F\subset A$ such that $A \subset \bigcup_\limits{x \in F} B(x, \varepsilon) $.

This is part of a bigger problem I want to prove.

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Figurinha
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    $A \subset \bigcup_{x\in F}B(x,\varepsilon/2) \implies \overline{A}\subset \bigcup_{x\in F}\overline{B(x,\varepsilon/2)} \subset \bigcup_{x\in F}B(x,\varepsilon)$. – Vinícius Novelli Oct 27 '17 at 20:51
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    This is also the $\Longrightarrow$ direction of https://math.stackexchange.com/questions/2958346/how-to-show-that-a-subset-x-is-totally-bounded-iff-bara-is-totally-bo?rq=1 . – darij grinberg Sep 19 '19 at 16:32

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For completeness, let me post an answer to conclude the question.

First of all, we want to prove that for any $\varepsilon>0,$ there exists a finite set $F\subseteq \overline{A}$ such that $$\overline{A}\subseteq\bigcup_{x\in F}B(x,\varepsilon).$$ Since $A$ is totally bounded, there exists a finite set $F\subseteq A\subseteq \overline{A}$ such that $$A\subseteq\bigcup_{x\in F} B(x,\varepsilon/2).$$ Observe that $$\overline{A}\subseteq\bigcup_{x\in F}\overline{B(x,\varepsilon/2)}\subseteq \bigcup_{x\in F}B(x,\varepsilon).$$ Therefore, $\overline{A}$ is totally bounded.

Idonknow
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