TITLE: Product of two compact spaces is compact
QUESTION [11 upvotes]: I read the proof that uses tube lemma and I do not have any problem with it but I cannot see what is wrong with the proof that first came to my mind: 
Let $X$ and $Y$ be compact spaces. Let $\mathcal{A}$ be an open covering of $X\times Y$. Then, $\bigcup_{U\in \mathcal{A}}{U}=X\times Y$. And $\pi_1(\bigcup_{U\in\mathcal{A}}{U})=\bigcup_{U\in\mathcal{A}}{\pi_1(U)}=X$, similarly $\bigcup_{U\in\mathcal{A}}{\pi_2(U)}=Y$. Since $X$ is compact, there are finitely many $\pi_1(U)$'s and $\pi_2(U)$ that cover $X$ and $Y$. So the product of these finitely many sets covers $X\times Y$. 
I presume that there is a mistake related to basic set theory.

REPLY [4 votes]: You find that the product of a finite number of those projected open sets covers your space, but what you have found is not a subcover.