TITLE: Cartesian product of compact sets is compact
QUESTION [21 upvotes]: Prove that if two sets $A$ and $B$ are compact then so is their Cartesian product $A \times B = \{(a,b): a \in A, b\in B\}$. 
The hint is to use Bolzano Weiertrass theorem and an argument of sequence to proof the statement.

REPLY [49 votes]: There is indeed a topological proof using the open covers definition of compactness.
Let $A$ and $B$ be compact sets and $\{O_\lambda\}_{\lambda\in\Lambda}$ be an open cover of $A \times B$. For each $(a,b) \in A \times B$, we can choose some $\lambda = \lambda(a,b)$ such that $(a,b) \in O_{\lambda(a,b)}$. By construction, $O_{\lambda(a,b)}$ is open, hence the point $(a,b)$ is contained in some open box $X \subset O_{\lambda(a,b)}$ where $X = U_{(a,b)} \times V_{(a,b)}$, where $U_{(a,b)} \subset A$ and $V_{(a,b)} \subset B$.
Suppose we fix $a$ and vary $b$. Then for every point $(a,b)$ we find that the point is contained in an open box in the product $A \times B$, and that box is then itself the product of a subset of $A$ with a subset of $B$. Proceeding in this manner, we observe that the collection of sets $\{V_{(a,b)}\}_{b\in B}$ is an open cover of $B$. Since by assumption $B$ is compact, we can find a finite cover $\{V_{(a,b_j(a))}\}$ of $B$ that consists of finitely many open sets containing points $\{(a,b_j(a))\}$.
Now let $U_a = \bigcap_j U_{(a,b_j(a))}$. Since $U_a$ is the intersection of finitely many open sets, it is itself open. Since $A$ is compact, there are finitely many $a_i$ such that $\{U_{a_i}\}$ forms an open cover of $A$. Then it follows that the collection of sets $\{O_{(a_i,b_j(a_i))}\}$ (for all combinations of $i,j$) is a finite cover of $A \times B$, hence $A \times B$ is compact.