Heine-Borel theorem. I'm interested in one question about Heine-Borel theorem.
We know that if S is bounded and closed then it's compact.
Standard proof using fact that some segment is compact. And because of S is bounded it's have left and right bound of segment. But my question is my S should be closed? We could take open , bounded S and some segment which cover out set. Why it's so necessary ?

An open interval, say $(0,2)$ is not compact in the reals, while $[0,2]$ is. 
Generally a subset of a compact set is not compact. 
Note that while one characterization of compact is that every sequence in $C$ has a convergent subsequence this means it has to converge in $C$, that is the limit must be in $C$. 
So for example the sequence $(1/n)_{n\ge 1}$ shows that the set $(0,2)$ is not compact. The sequence does not converge as a sequence in $(0,2)$ since its limit (in the reals) is outside the set. 
If you consider the definition via open coverings the point is that a smaller will have coverings that won't covert the larger and those might not allow a finite subcover. 
Also this may be counter-intuitive at first,  but covering the larger sets might give you a "good" building block for a cover that allows you to ditch many other sets from a cover of the smaller set. 
Take the example basically as in the  other answer $(1/n,2)$ for $n\ge 1$ which is an open cover of $(0,1]$ with no finite subcover. But if you were to cover the point $0$ too (so the compact interval $[0,1]$), you'd need a "new" open set and this open set will allow you to discard almost all of the former sets. Say you cover $0$ with $(-0.001, 0.001)$ then you only need the former sets for $n$ up to $1001$  to cover $[0,1]$, so you have a finite cover. If you cover $0$ with something still smaller just increase the $n$ but always a finite number suffices as you'll cover $0$ with an open set that thus contains some interval $(-\epsilon, \epsilon)$.
