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.

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 $S$.
