Closed subset of compact set is compact 
If S is a compact subset of R and T is a closed subset of S,then T is compact.
(a) Prove this using definition of compactness.
(b) Prove this using the Heine-Borel theorem.

My solution: firstly I should suppose a open cover of T, and I still need to think of the
set S-T. But if S-T is open in R,it can be done because the open cover of T and S-T is a open cover of R. The reality is S-T is not necessarily a open set in R. My question is that How we can find a open cover which covers S-T but misses T! I don't know how to do this thing!
in terms of part (b), I know it is bounded, but How to prove it is closed for T.
 A: Here's how to get started on part a. 
Start with an open cover of $T$. You need to show it has a finite subcover. If $U$ is in the open cover, then it's open in $T$, which means that there's an open set $U'$ in $S$ such that $U = T \cap U'$. For every $U$ in your cover, find a corresponding $U'$; now you almost have a cover of $S$. As you observed, if you add in $S - T$, you now have an open cover of $S$. What do you know about every open cover of $S$? 
A: $T$ is a closed subset of $S$ if and only if $T=C\cap S$ for some $C$ closed in $\mathbb{R}$. But $S$ is closed too, being compact, so $T$ is closed in $\mathbb{R}$ because it is the intersection of two closed sets. This takes care of the remaining part of $(b)$. For $(a)$, $\mathbb{R}\setminus T$ is an open set containing $S$.
A: I will address part (b), since the others address (a) well.
The Heine-Borel theorem says that a subset $V \subset \mathbb{R}$ is compact if and only if it is both closed and bounded.
So suppose that $T \subset S \subset \mathbb{R}$. Since $S$ is compact, it is closed and bounded. What can you now say about the boundedness and closedness of $T$?
