$(M,d)$ is complete iff all closed and countable subspace of $M$ is complete

Question

A metric space $(M,d)$ is complete $\iff$ if every closed and countable subspace $F\subseteq M$ is complete.

$\implies)$ For this implication I use a proposition that says: "If $X$ is a complete metric space and $F \subseteq X$ is closed, then $F$ is complete". But for $\impliedby )$ I don't know what to do.

Answer 1

As suggested by Daniel Fischer, consider the set $A=\{a_1,a_2,\ldots\}$, where $\{a_i\}_{i>0}$ is a Cauchy sequence. A Cauchy sequence has at most one limit point, so the closure $\overline{A}$ is still countable as it contains at most one extra point. By hypothesis, $\overline{A}$ is complete because it is a closed countable subset, hence the Cauchy sequence converges to a limit $a\in\overline{A}$. Thus the Cauchy sequence converges to $a$ in $M$ as well. Since this is true for any Cauchy sequence, $M$ is complete.