TITLE: Proving that the Union of Two Compact Sets is Compact
QUESTION [13 upvotes]: Prove if $S_1,S_2$ are compact, then their union $S_1\cup S_2$ is compact as well.

The attempt at a proof:
Since $S_1$ and $S_2$ are compact, every open cover contains a finite subcover. Let the open cover of $S_1$ and $S_2$ be $\mathscr{F}_1$ and $\mathscr{F}_2$, and let the finite subcover of $\mathscr{F}_1$ and $\mathscr{F}_2$ be $\mathscr{G}_1$ and $\mathscr{G}_2$, respectively. If I can show that $S_1\cup S_2$ contains a finite subcover for every open cover, then I will have showed that the union is indeed compact. I note that $\mathscr{G}_1\subset\mathscr{F}_1$ and that $\mathscr{G}_2\subset\mathscr{F}_2$ (by definition of an open subcover). Then, I note that $\mathscr{G}_1\subset\mathscr{F}_1\cup\mathscr{F}_2$ and that $\mathscr{G}_2\subset\mathscr{F}_1\cup\mathscr{F}_2.$
I'm not sure how to proceed from this point. I think I am on the right track though. Any suggestions to proceed would be appreciated. Thanks in advance.

REPLY [6 votes]: Every open cover of the union is an open cover of each set.
So for each set there is a finite subcover.
The union of the finite subcover is still finite and covers the union of the two sets.
So the union is indeed compact.