Topological proof of the compactness of product metric space 
Problem. Let $(X,d_X)$ and $(Y,d_Y)$ be two compact metric spaces (see the definition here). Then show that the product metric space $(X\times Y,d_{X\times Y})$ is also compact.

Now this can be done easily using sequences but the proof using the notion of sequences doesn't seem to be general enough. So I was looking for a "purely topological" proof but can't find one. 
Can anyone help?
 A: Take a cover by open sets $U_\alpha$. 
First fix $x \in X$. The set $\{x\} \times Y$ is homeomorphic to $Y$ and so compact. It is covered by the $U_\alpha$ and so there is a finite subcover $U^x_{\alpha_1}, \dots, U^x_{\alpha_{n_x}}$ of $\{x\} \times Y$. Let $U_x := \bigcup^{n_x}_{i=1} U^x_{\alpha_i}$. Since $U_x$ is open, for each $y \in Y$ there is a basic open set $V^x_y \times W^x_y$ contained in $U_x$ and containing $(x,y)$. Since $Y$ is compact and the sets $\{W^x_y : y \in Y\}$ cover $Y$, there is a finite set $F^x \subseteq Y$ such that $\{W^x_y : y \in F^y\}$ covers $Y$. Let $V^x := \bigcap_{y \in F^x} V^x_y$. Then the set $V^x \times Y$ is contained in $U_x$.
Now since $X$ is compact, the cover $\{V^x : x \in X\}$ of $X$ by open sets has a finite subcover $V^{x_1}, \dots, V^{x_k}$. It follows that the sets $V^{x_1} \times Y, \dots, V^{x_k} \times Y$ cover $X \times Y$ and so the sets $U_{x_1}, \dots, U_{x_k}$ cover $X \times Y$. Hence
$$
\{U^{x_j}_{\alpha_i} : j \le k, i \le n_{x_j}\}
$$
is a finite subcover of the original cover $U_\alpha$.
