(Proof Checking) Prove that if $C,C'$ are compact subsets of a hausdorff space $X$ then $C\cap C'$ is compact in $X$. Prove that if $C,C'$ are compact subsets of a hausdorff space $X$ then $C\cap C'$ is compact in $X$.
I am tempted to use the following argument. Let $U = \{U_i|i\in I\}$ be some open cover of $C\cap C'$ where $I$ is a possibly infinite indexing set. Also, let $V= \{ V_j | j\in J   \}$ be an open cover for $C$ for some indexing set $J$. Now let $W$ be the sum of $U$ and $V$, that is, $W$ contains all open sets in $U$ and $V$. Then $W$ is an open cover for $C$. Because $C$ is compact we may find a finite subcover for $C$, call it $W_f$. Then we remove from $W_f$ all sets that were not originally in $U$. We are left with an open cover for $C\cap C'$ that is finite, and is a subcover of $U$. Therefore $C\cap C'$ is compact.
I feel I made a mistake somewhere because I did not use the Hausdorff condition. Can anyone point out the mistake and possible provide a hint to a proof that is correct? Thanks advance!

Since $C$, $C'$ are compact and $X$ is a hausdorff space, they are closed. Therefore $C\cap C'$ is closed. Now, closed subsets of compact sets are compact. 
