A closed subset of a complete metric space is a complete subspace. Proof. Let $S$ be a closed subspace of a complete metric space X. Let $(x_n)$ be a Cauchy sequence in $S$. Then $(x_n)$ is a Cauchy sequence in $X$ and hence it must converge to a point $x$ in $X$. But then $x \in \bar{S} = S$. Thus $S$ is complete.
I have seen this theorem in several places but I never know why they are able to say "But then $x \in \bar{S} = S$"...Why is $x$ in $\bar{S}$?