Csdrhrt

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.

Assuming that your space is Hausdorff $Ksetminus E_2=Kcap E_1$ and $Ksetminus E_2$ is compact. On second thoughts you don't need Hausdorff property!.

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 $Kcap E_1 = K cap (Xsetminus E_2)$ and thus is a closed subset of the compact $K$ and hence compact too.