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.
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.