Proof of Compactness. I am reviewing for my topology final and I came across this problem that I can't figure out.  
Use the definition of compactness to prove that if $A$ is closed in $X$ and $B$ is compact in $X$ then $A \cap B$ is compact in $X$.
I know that the cover of $A$ can be written as open sets from $X$ and I assume we need the fact that $X-A$ is open add it to the cover and then come up with the appropriate finite sub-cover but I can't seam to get anywhere. 
Thank you.
 A: You almost have it, but you're covering the wrong set.
Suppose $\mathcal C$ is a cover of $A\cap B$ by open subsets of $X$. Then $\mathcal C\cup\{X\setminus A\}$ covers $B$. Select a finite subcover from that (because you know $B$ is compact), and remove $X\setminus A$ if it's in it. The resulting finite subfamily of $\mathcal C$ must cover $A\cap B$.
A: Let $U_i, i\in I$ be an open cover of $A\cap B$, you can write $U_i=V_i\cap A\cap B$ where $V_i$ is an open subset of $X$. $(V_i\cap B, (X-A\cap B)\cap B)$ is an open cover of $B$, since $B$ is compact, you can extract a finite cover, $V_1\cap B,...,V_n\cap B, (X-A\cap B)\cap B$. $U_1,..,U_n$ is an open cover of $A\cap B$. 
A: Say $\mathcal{U}$ is an arbitrary open cover of $A\cap B$. The coplement of $A\cap B$ in $B$ (say C) is open according the subspace topology of $B$. So we can write C=B$\cap$O for an open set O. So $\mathcal{U}\ \cup$ {O} is an open cover of $B$ .  Since $B$ is compact we have a finite subcover $\mathcal{U'}$ of it. So  $\mathcal{U'}$ is also finite subcover of $A\cap B$.
