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?
 A: 
How do I know that there is not an open cover that covers $F$ but not $K$?

You do not know. For example if $F = K$, then every open cover of $F$ also covers $K$. But this is none of your business. 
I guess I understand your confusion now. Let me spell out the proof.
Your assumption is that $\{ V_\alpha\}_\alpha$ covers $F$. So $\{ V_\alpha\}_\alpha \cup F^c$ covers $K$. From there you can use the assumption that $K$ is compact and pick a finite subcover. The finite subcover looks like 
$$ \{ V_{\alpha_1}, V_{\alpha_2} , \cdots ,V_{\alpha_k}, F^c\}.$$
This covers $K$ and in particular covers $F$. But now we can take $F^c$ away: we know by definition of $F^c$ that $F\cap F^c$ is empty anyway. Thus 
$$ \{ V_{\alpha_1}, V_{\alpha_2} , \cdots ,V_{\alpha_k}\}$$
covers $F$ and is a finite subcover of $\{V_\alpha\}_\alpha$. 
A: 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. 
A: 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.
