Proof. The proof is by the straightforward strategy of composing a finite subcover from a lower-dimensional subcover. Let the open cover $\mathcal{C}$ of $X \times Y$ by basis sets be given.

The set $X \times \{ y \}$ is compact, because it is the image of a continuous embedding of the compact set $X$ . Hence $X \times \{ y \}$ has a finite subcover in $\mathcal{C}$ : label the subcover by . Do this for each $y \in Y$ .

To get the desired subcover of $X \times Y$ , we need to pick a finite number of $y \in Y$ . Consider $V^y = \bigcap_{i=1}^{k_y} V_i^y$ . This is a finite intersection of open sets, so $V^y$ is open in $Y$ . The collection $\{ V^y : y \in Y\}$ is an open covering of $Y$ , so pick a finite subcover . Then $\bigcup_{j=1}^l \mathcal{S}^{y_j}$ is a finite subcover of $X \times Y$ .