If $A \subseteq B_1 \cup B_2$ where $B_1$ and $B_2$ are disjoint open sets and $A$ is compact, show that $A \cap B_1$ is compact. 
If $A \subseteq B_1 \cup B_2$ where $B_1$ and $B_2$ are disjoint open sets and $A$ is compact, show that $A \cap B_1$ is compact.

I tried to prove it : 
Suppose that $A$ is compact and $A \subseteq B_1 \cup B_2$ where $B_1$ and $B_2$ are open and disjoint. Let $X = \displaystyle \bigcup_{\omega \in \Omega} V_{\omega}$ be an open cover of $A \cap B_1$. Then $A \subseteq \left(A \cap B_1 \right) \cup B_2$, implying that $X$ and $B_2$ cover $A$. Since $A$ is compact, the cover $X$ has a finite subcover $X^\prime= \bigcup_{i=1}^n U_i$, so we can write $$A \cap B_1 \subseteq A \subseteq X^\prime \cup B_2,$$ implying that $ A \cap B_1 \subseteq X^\prime$ since $B_1 \cap B_2 = \emptyset$. Thus $A \cap B_1$ is compact. $\Box$
Does my proof work?
 A: The basic idea is correct, but there are a few problems. First, your $X$ isn’t a cover of anything: a cover is a family of subsets of the space, and your $X$ is subset, not a set of subsets. What you mean is that if $\mathscr{V}=\{V_\omega:\omega\in\Omega\}$ is some open cover of $A\cap B_1$, then $\mathscr{V}\cup\{B_2\}$ is an open cover of $A$. Your $X$ is the union of the cover $\mathscr{V}$: $X=\bigcup\mathscr{V}$. 
Next, compactness of $A$ doesn’t tell you that $\mathscr{V}$ has a finite subcover: it tells you that $\mathscr{V}\cup\{B_2\}$ has a finite subcover, say $\{U_1,\ldots,U_n\}$. If $U_k\in\mathscr{V}$ for $k=1,\ldots,n$, we’re done: $\{U_1,\ldots,U_n\}$ is a finite subset of $\mathscr{V}$ that covers $A\cap B_1$. If not, one of the $U_k$ must be $B_2$. Without loss of generality we can assume that it’s $U_n$, and since $B_2\cap(A\cap B_1)\subseteq B_2\cap B_1=\varnothing$, we conclude that $\{U_1,\ldots,U_{n-1}\}$ is a finite subset of $\mathscr{V}$ covering $A\cap B_1$.
I’m sure that this is what you had in mind; it just isn’t quite what you wrote.
