Question on the proof that a closed subset of a compact set is compact I was reading the proof here on the claim that a closed subset of a compact set is compact, which reads:

Say F ⊂ K ⊂ X where F is closed and K is compact. Let $\{V_α\}$ be an open
  cover of F. Then $F^c$ is a trivial open cover of $F^c$. Consequently {
  $F^c$} ∪ $\{V_α\}$ is an open cover of K. By compactness of K it has a finite
  sub-cover – which gives us a finite sub-cover of F.

The proof has to add $\{F^c\}$ to an open cover of of F so that it covers K. What about the open cover $\{V_α\}$ without $\{F^c\}$? How can one be sure that it has a finite subcover for F since it is not necessarily true that $\{V_α\}$ covers K?
UPDATE
I want to put the question differently. How do I know that there is not an open cover that covers F but not K?

You don't have to know that $V_{alpha}$ doesn't cover $K$, if it happens to cover $K$ then still adding $F^C$ wouldn't alter anything - the resulting cover would still cover $K$.
Now of course in the subcover of $K$ including $F^C$ we must know that the subcover excluding $F^C$ would cover $F^C$, but this is rather obvious. If $x\in F$ then it would be in at least one of the set in the subcover, but it can't be in $F^C$ by definition so it must be in at least another of the sets in the subcover. 
