Understanding a proof in Rudin 
Closed subsets of compact sets are compact
Proof: Suppose $F \subset K \subset X$, $F$ is closed (relative to $X$), and $K$ is compact. Let $\{V_{\alpha}\}$ be an open cover of $F$. If $F^c$ is adjoined to $\{V_{\alpha}\}$, we obtain an open cover $\Omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers $K$, and hence $F$. IF $F^c$ is a member of $\Phi$. we may remove it from $\Phi$ and still retain an open cover of $F$. We have thus shown that a finite sub collection of  $\{V_{\alpha}\}$ covers $F$

So the second step confuses me. Basically they are saying the union of the complment of $F$ and the open cover of $F$ covers $K$. I don't see why this is true. Why can't $F^c \subset K$? If that were true, doesn't it mean it wouldn't necessary cover $K$?
 A: $\forall x\in K, (\text{and}\;\;\forall x \in X)$: $x\in F\cup F^c$. $\;x\in F$ or $x\in F^c \implies (x \in F$ or $x \notin F$). That is $x$ is in $F$, or it's not in $F$ (Law of excluded middle).   If it's in $F$, it's covered by the open cover of $F$: $\{V_\alpha\}$.  If it's in $F^c$ (i.e., if it's not in $F$), it's covered by $F^c$.  
Whatever the case, it's covered: all of $X$ so all of $K$ is covered, and since K is compact, it can be covered by a finite subcover, and following the proof, $F$ can then be covered by a finite collection of $\{V_\alpha\}$, so it too is compact.

Edit: Since the problem seems to be in understanding $F^c$, the complement of $F$ relative to $X$:
Complement clarification: See image.
Let $X = U = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. 
Let $F = B = \{0, 1, 4, 5, 6, 7, 8\}.$ 
Then $F^c = B'= X \setminus F = U \setminus B = \{2, 3, 9\}$

A: Every member of $K$ is either in $F$ or in $F^c$.  If it's in $F$, it's covered by the open cover of $F$.  If it's in $F^c$, it's covered by $F^c$.  So in either case, it's covered.  
A: $$X\subseteq F\cup F^c$$
Since $F\subseteq\cup\Omega$ (Here $\cup\Omega$ is the union of all elements of $\Omega$), therefore:
$$X\subseteq\cup\Omega\cup F^c$$
Thus:
$$K\subseteq\cup\Omega\cup F^c$$
A: The open cover you're dealing with is $\{F^c\} \cup \{V_\alpha\}$.  This covers the entire space $X$, not just $K$. To see this, let $x \in X$.  Either $x \in F$ or $x \notin F$.  If $ x \in F$, then
$$x \in F \subseteq \bigcup V_{\alpha}$$
Otherwise, $x \in F^c$, so $X \subseteq F^c \cup \bigcup V_{\alpha}$.
