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Let $(X,d)$ be complete, $S\subseteq X$.

Want to show that $(S, d|_{S\times S})$ is complete if and only if $S$ is closed in $(X,d)$.

Can anyone give me any hints? Thank you all.

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    If $S$ is complete consider a limit point of $S$ and a sequence of points from $S$ that converge to it. If $S$ is closed, consider a Cauchy sequence in $S$ and the point in $X$ that it converges to. – 2017-01-27
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    @chilangoincomprendido Assume $S$ is complete, then $\exists (x_n)$ Cauchy s.t. $(x_n)\rightarrow k\in X$. But I did not see the connection. – 2017-01-27

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Suppose $S$ is complete. Assume $S$ is not closed, then $Clsr(S)\setminus S\neq\emptyset$. Choose $x\in Clsr(S)\setminus S\neq\emptyset$, then for every $n\in\mathbb{N}$, we have $(B(x,\frac{1}{n})\setminus\{x\})\cap S\neq \emptyset$, so we can choose $x_n\in (B(x,\frac{1}{n})\setminus\{x\})\cap S$. Obviously $(x_n)$ is a Cauchy sequence in $S$ with $x_n\to x$, but $x\notin S$ which means $S$ is not complete, contradiction.

Conversely suppose $S$ is closed. Assume $S$ is not complete. Let $(x_n)$ be a Cauchy sequence in $S$ which does not converge in $S$. Then since $X$ is complete, there exists $x\in X\setminus S$ such that $x_n\to x$. This means for every $\epsilon>0$, $B(x,\epsilon)$ contains a tail of the sequence, and thus has non-empty intersection with $S$. This implies $x\in Clsr(S)$, and therefore $S\neq Clsr(S)$ and this means $S$ is not closed, contradiction.