Closed subset of complete metric space...don't understand last part of theorem. 
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 $\bar{S}$?
 A: Let $U$ be any open nbhd of $x$. There is an $r>0$ such that $B(x,r)\subseteq U$, where $B(x,r)$ is the open ball of radius $r$ centred at $x$. Since $\langle x_n:n\in\Bbb N\rangle\to x$, there is an $m\in\Bbb N$ such that $d(x_n,x)<r$ whenever $n\ge m$. In particular, $d(x_m,x)<r$, so $x_m\in B(x,r)\subseteq U$. Moreover, $x_m\in S$, so $x_m\in U\cap S$. This shows that $U\cap S\ne\varnothing$ for every open nbhd $U$ of $x$, and it follows at once that $x\in\operatorname{cl}S$.
Added: Here in one place are some definitions that seem to be giving you a bit of trouble. Let $\langle X,d\rangle$ be a metric space, $S\subseteq X$ and $x\in X$.


*

*$x\in\operatorname{cl}S$: for each $r>0$, $B(x,r)\cap S\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$.  

*$x$ is a limit point (or cluster point) of $S$: for each $r>0$, $B(x,r)\cap(S\setminus\{x\})\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$ other than $x$.  

*$x\in\operatorname{int}S$: there is an $r>0$ such that $B(x,r)\subseteq S$.

