A subset of a metric space is closed iff it contains all of its cluster points. A point, $p$, is defined as a cluster point of a set $S$ if $\forall \epsilon > 0,$ there exists an open ball of radius $\epsilon$ centered at $p$ that contains infinitely many points in $S$.
We want to prove that a subset $S$ of a metric space $(E,d)$ is closed iff $S$ contains all of its cluster points.
I believe I have the forward direction, which seems to be pretty straightforward, but I am having difficulty with the $(\Leftarrow)$ direction. My idea was to somehow use the fact that $S$ contains all of its cluster points to show that every infinite subset of $S$ contains a cluster point. Then we can say $S$ is sequentially compact, therefore it is compact, and hence closed.
I'm just mostly unsure of how to show the first part. Thanks for any comments.
 A: Suppose that $S$ is not closed. The $S\varsubsetneq\overline S$. Take $x\in\overline S\setminus S$. Then $x$ is a cluster point of $S$ that doesn't belong to $S$.
A: Given the definition $S\subset E$ is closed iff the complement of S, $E\setminus S$ is open. 
A set is Open if all of its points are Interior Points. A point $x\in S$ is an Interior point of S if   $\exists \epsilon >0 $ such that if $d(x,y)<\epsilon$, then $y\in S$. In other words an open ball of radius $\epsilon$ centered at $x$ contains only elements of $S$. So that open ball is itself a subset of $S$ completely contained in $S$. 
Suppose a set $S$ contains its cluster points. Let $x\in S$ be a cluster point. Then $\forall \epsilon>0$, an open ball of radius $\epsilon$ centered at $x$ contains infinitely many points of $S$. Let $y\in E\setminus S$ and suppose $\forall \delta>0$, an open ball centered at $y$ of radius $\delta$ contained some point of S. For any such point, reduce $\delta$ to be that new point's distance from y. This generates an infinite of points, so $y$ is a cluster point of $S$, and so must be contained in $S$. This contradicts our hypothesis that y is in $E \setminus S$. It follows that $\exists \delta$ such that an open ball centered at y contains only points in $E\setminus S$, so $y$ is an interior point of $E\setminus S$. By the above, such a $\delta$ must exist for every $y$, so every $y$ is an interior point of $E\setminus S$. Since every point of the complement is an interior point, he complement is Open and therefor S is closed. 
Other Direction:
Suppose $S$ is closed. Then $E\setminus S$ is Open. Let $x\in S$ be a cluster point of $S$. By definition of a cluster point, every open ball centered at $x$ must contain an infinite number of points of S. If $x\in E\setminus S$, $x$ has to be an Interior Point of $E\setminus S$ since it is open. But by definition, an interior point must have some open ball centered on it containing no elements of the complement. No cluster point of S can be an interior point of $E\setminus S$, so it must not be in $E \setminus S$. It is therefor in S, thus S contains all of its cluster points.  
