Q:

Prove Heine-Borel Theorem

Prove Heine-Borel Theorem: "A subset $S$ of $\mathbb{R}$ is compact if and only if every open cover for $S$ has a finite subcover."
Suggestions: Let $S \subset \mathbb{R}$. If every open cover for $S$ has a finite subcover, then $S$ must be compact. Why? Now, assume that $S$ is compact and let $\mathcal O$ be any open cover for $S$. Let $\epsilon > 0$ so that for all x in $S$, there is come $E \in \mathcal O$ such that $D(x,\epsilon) \subset E$. Assume that S is totally bounded.

A:

This is too long for a comment, so I'll make it an answer.
I'm not sure if I understand your question.  There seems to be a typo.  As you know, it is not the case that every open cover of $\mathbb{R}$ has a finite subcover.  If $S$ is unbounded, it is easy to find an open cover (that covers $\mathbb{R}$, not just $S$) with no finite subcollcection covering $S$.  If $S$ is not closed, $S$ has a limit point $x$ that doesn't belong to $S$, and it is easy to construct an open cover of $\mathbb{R} \setminus \{x\}$ (which contains $S$) with no finite ssubcollection covering $S$.  
The existence of $\epsilon$ that you are using in your argument is far from obvious (since $\epsilon$ does not depend on $x$), though such an $\epsilon$ does exist. Are you allowed to assume this $\epsilon$ exists, or do you need to prove it?  You only need it if you are going to use the idea of sequential compactness, which is not necessary for what you seem to be trying to prove.  
Proving the other direction, that any closed and bounded subset of $\mathbb{R}$ is compact, is more difficult, but not terribly difficult.  You can probably Google it and find a complete, excellent proof faster than you will get one here.  You will also find proofs of the easier direction, as I describe above.  It would also not surprise me if this question and been asked and answered in this forum before.

