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?

The idea of the proof is that, according to the definition of compactness, every open cover for $K$ has a finitie subcover. Now we found an open cover for $K$, containing only one set more than the open cover for $F$, we conclude that this set has a finite subcover. Since $F^c \not\subset F$, we know that there exists a subset of the finite subcover for $K$ also covering $F$, that does not contain $F^c$, and is finite since it contains as most as many sets as the finite cover for $K$.
To answer your update, there exist open covers for $F$ not covering $K$, but by construction, adding $F^c$ makes suren the open cover also covers $K$. That makes the possibility you're asking about irrelevant.
