Compactness of intersection of a compact set and an open set

Question

If $K \subset E_1 \cup E_2$, where $K$ is compact and $E_1, E_2$ are disjoint open subsets of a topological space, is $K \cap E_1$ compact? Is that always the case if $E_1, E_2$ are not disjoint?

I've seen other threads that have this, so I was wondering why the solution I thought of is incorrect:

Let $U_{\alpha}$ be an open covering of $K$. Then because $K$ is compact, there is a finite subcovering $U_1, U_2, \ldots, U_N \in U_{\alpha}$ that cover $K$. But then $U_1, \ldots, U_N, E_1$ is a finite collection of open sets that covers $K$ and $E_1$, so it covers $K \cap E_1$, and so $K \cap E_1$ must be compact.

Answer 1

Assuming that your space is Hausdorff $K\setminus E_2=K\cap E_1$ and $K\setminus E_2$ is compact. On second thoughts you don't need Hausdorff property!.

Answer 2

Your proof should start with a cover $\mathcal{U}$ (WLOG by open sets of $X$) of $K \cap E_1$, and produce a finite subcover of that:

Add the one set $E_2$ to $\mathcal{U}$ and we have an open cover of $K$ (any set in $K$ lies in $E_1$ or $E_2$ and the first ones are covered by $\mathcal{U}$ the other by $E_2$..) and so $\mathcal{U} \cup \{E_2\}$ has a finite subcover $\mathcal{U}'$ and then $\mathcal{U}'\setminus \{E_2\}$ is still finite (smaller than the finite $\mathcal{U}'$) and a subcover of $\mathcal{U}$. So $K \cap E_1$ is compact.

Alternatively note that $K\cap E_1 = K \cap (X\setminus E_2)$ and thus is a closed subset of the compact $K$ and hence compact too.