Show the closure is compact 
Let $E$ be a bounded subset of $\mathbb{R}$. Prove that $\overline{E}$ is compact.

I am trying to prove this, but I keep getting stuck. I know that if $E$ is closed and bounded then I can conclude that $E$ is compact. I also know that $\overline{E}$ is closed, but I need to show $E$ is closed. However, I don't understand how I would do this. Could someone please demonstrate how to show $E$ is closed. 
Note(Definition): If $E$ is a subset of $\mathbb{R}$, let $E'$ denote the set of limit points of $E$. The closure of $E$, denoted  $\overline{E}$ is defined as $\overline{E}= E \cup E'$. 
 A: You do not need to show that $E$ is closed (as what @)carmichael561 said in his comment.)
Well, $E$ is assumed to be bounded and so, we can find a closed interval $I$ such that $E\subset I$. This implies that $I$ is a closed set that contains $E$. But $\overline{E}$ is the smallest closed set that contains $E$. Thus, $\overline{E}\subset I$. This implies that $\overline{E}$ is bounded. Since $\overline{E}$ is also closed, it follows from the Heine-Borel Theorem that $\overline{E}$ is compact.
A: " but I need to show E is closed."
As $E$ can be any arbitrary bounded set you can not prove it is closed as it might not be.  (Say $E = (0,5)$).
From your comment, you are hoping if $E$ closed then $\overline E = E$ and so $\overline E$ is bounded.  That is ... hope.  It's akin to hoping to prove that the algebraic numbers are countable, you hope to prove the algebraic numbers are rational.  There's no reason to assume they are and if given nothing else we can't.
As sets of $\mathbb R^n$ are compact if and only if they are closed and bounded.  So you need to prove.
A) If $E\subset \mathbb R$ then $\overline{E}$ is closed.
Hint: Prove if $p$ is a limit point $\overline {E}$ then $p$ is a limit point of $E$.  I.e.  if every neighborhood of $p$ contains a point of $\overline{E}$ then every neighborhood of $p$ contains a point of $E$.
Hint 2: To prove that, prove if $p' \in B_{\epsilon}(p)$ is a limit point of $E$ then there is a neighborhood $B_{\epsilon'}(p') \in B_{\epsilon}(p)$.
Hint 3: Hint 2 has nothing to do with the set $E$ or limit points.  Given any point $r \in \mathbb R^n$ and any $\epsilon > 0$ and any $s \in B_{\epsilon}(r)$ then will be somoe $B_{\epsilon'}(s) \subset B_{\epsilon}$.
B) prove that if $E$ is bounded prove $\overline E$ is also bounded.
This is really the heart of the problem and not hard at all.
Hint:  If $|x| \le M$ for all $x \in E$, and if $|p-x| < \epsilon$ then prove $|p| < M + \epsilon$.  
