Prove that if $K$ and $L$ are compact subsets of a Hausdorff space $X$, then $K \cap L$ is a compact subset of $X$.
I understand that since $K$ and $L$ are compact subsets, they each have finite coverings. Do I have to show that the intersection of $2$ finite coverings is a compact subset?
You can find a proof here.
Let $\mathcal{A}$ be an open cover of $K \cap L$. Since $K$ and $L$ are closed sets ,$K \cap L$ is also closed. So ${(K \cap L)}^C$ is open. Then $\mathcal{B}=\mathcal{A}\cup {(K \cap L)}^C$ is an open cover of $K$(and $L$). Since $K$ is compact,$\mathcal{B}$ has a finite subcover,say,$\{A_1,A_2,\ldots,A_n,{(K \cap L)}^C\}$. So $\{A_1,A_2,\ldots,A_n\}$ is a finite subcover of $\mathcal{A}$ that covers $K \cap L$. Hence, $K \cap L$ is compact.