Direct proof for closure of a set $E$ is closed. Let $X$ be a metric space and $E\subset X $ be a subset of $X$.   Let $cl(E)=E\cup E'$, where $E'=$ set of all limit points of $E$. 
Claim: $cl (E)$ is closed. 
If $E$ is finite, then E has no limit points and therefore $cl (E)= E$ and since E contains all its limit points (which are none) so vacuously $E$ is closed.
So let $E$ be infinite.
Proof: Let $x\in X$ be a limit point of $cl (E)$. Then given any $\delta\gt 0, B_X(x,\delta)\cap cl (E)$ is infinite.  It follows that either $B_X(x,\delta)\cap E$ or $B_X(x,\delta)\cap E'$ is infinite. This implies that either $x\in E'$ or $x\in (E')'$. (?) 
Since, $E'$ is a closed set, it follows that $(E')'\subset E'$ and therefore $x\in E'$ or $x\in (E')'\implies x\in E'\subset cl(E)$. Since $x$ is arbitrary, it is proven that $cl(E)$ is closed.
Is my proof correct? Thanks. 
Edit: Closed set definition being used here: $A\subset X$, where $X$ is a metric space, is said to be closed if $A$ contains all its limit points.
 A: You can handle the finite & infinite cases together:
(1). For any $F\subset X$ we have $x\in F\cup F'$ iff every open set containing $x$ contains some member of $ F$ (which may or may not be equal to $x$).
(2). Let $x\in (E\cup E')'.$  Let $U$ be any  an open set containing $x$. By (1) with $F=(E\cup E')$, the set $U$ contains some $y\in (E\cup E')$. It does not matter whether or not $y=x$. What matters is that some $z\in E$ belongs to $U$ because
(i) if $y\in E,$ let $z=y,$ or
(ii) if $y\in E',$ then the open set $U,$ which contains $y,$  must contain some $z\in E.$
Hence by (i) & (ii) every open set $U$ containing $x$ must contain some $z\in E$. So $x\in (E\cup E')$ by (1) with $F=E$. So we have $$x\in (E\cup E')'\implies x\in (E\cup E').$$
A: Here is another approach:
Suppose $x \notin \operatorname{cl} E$, then $x \notin E$ and $x \notin E'$.
Then there must be a neighbourhood $U$ of $x$ that contains no point of $E$ (otherwise $x$ would be a limit point).
Furthermore, this holds for every element of $U$, hence $U$ does not intersect $\operatorname{cl} E$ and so we see that $(\operatorname{cl} E)^c$ is open from which we see that $\operatorname{cl} E$ is closed.
