Let $(X,d)$ be an arbitrary complete metric space and suppose $S\subseteq(X,d)$. Show that $S$ is closed if and only if every Cauchy sequence in $S$ converges to a point in $S$.
I did the forward direction, is it correct?
Suppose $S$ is closed. Let $x\in S$ be an Cauchy sequence, $x=(x_n)_{n \in \mathbb N}$. Then, since $S$ is closed, it contains all of its limit points. Therefore, $\lim_{n\to\infty} (x_n)$ will converge to an element in $S$. So every Cauchy sequence in $S$ converges to an element in $S$.
For the backward direction, would I just let $S\subseteq(X,d)$ and suppose that every Cauchy sequence in $S$ converges to a point in $S$. Then show that a limit point $x$ is in $S$?
Any feedback is appreciated, thanks.