the product of topological spaces is compact, then each of its factors is compact.

Solution 1:

You’re off on the wrong foot altogether. If you’re going to work directly with open covers (instead of @lhf’s idea), you must start with an open cover of $X$ (or $Y$) and show that it has a finite subcover. Starting with an open cover of $X\times Y$ is pointless: you’re not trying to prove that $X\times Y$ is compact. (And note that open sets in $X\times Y$ do not necessarily have the form $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$.)

Let $\mathscr{U}$ be an open cover of $X$. For each $U\in\mathscr{U}$ let $V_U=U\times Y$, and let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. Show that $\mathscr{V}$ is an open cover of $X\times Y$ and apply compactness of $X\times Y$ to get a finite subcover of $\mathscr{V}$. Use this finite subcover to find a finite subset of $\mathscr{U}$ that covers $X$.

The proof that $Y$ is compact is entirely similar.

Solution 2:

This answer expands on the brief comment of @lhf.

This is a standard theorem in point-set topology (which I won't prove here unless you insist).

Theorem: If $f:K \to L$ is a continuous map between topological spaces and $K$ is compact, then $L$ is compact as well.

And this is another standard proposition, where the product is given the product topology.

Proposition: The projection maps $\pi_i: X_1 \times X_2 \to X_i$ for $i = 1, 2$ are continuous.

Putting these two results together, you get your claim as a direct corollary.

Solution 3:

Your proof should start with an arbitrary open cover of $X$ (or $Y$) and then show the existence of a finite subcover (by using the compactness of $X\times Y$). Hint: If $U\subseteq X$ is open in $X$ then $U\times Y$ is open in $X\times Y$.