[proofplan] We prove $X \times Y$ is compact by the tube lemma technique. We may assume the open cover consists of basic open sets $U_\alpha \times V_\alpha$. For each $x \in X$, we cover the compact slice $\{x\} \times Y$ by finitely many basic rectangles, then intersect their $X$-projections to form an open tube $W_x \times Y$. The tubes $\{W_x\}_{x \in X}$ form an open cover of $X$, and compactness of $X$ yields a finite subcover, from which we assemble a finite subcover of $X \times Y$. [/proofplan] [step:Reduce to a cover by basic open sets] Let $\mathcal{U}$ be an open cover of $X \times Y$. Every [open set](/page/Open%20Set) in the product [topology](/page/Topology) is a union of basic open sets of the form $U \times V$ with $U$ open in $X$ and $V$ open in $Y$. Therefore we may refine $\mathcal{U}$ to a cover $\{U_\alpha \times V_\alpha\}_{\alpha \in A}$ by basic open sets: each $U_\alpha \times V_\alpha$ is contained in some member of $\mathcal{U}$. A finite subcover of the refinement yields a finite subcover of $\mathcal{U}$ (each basic rectangle is contained in a member of $\mathcal{U}$, so we select the corresponding members). [/step] [step:Cover each compact slice $\{x\} \times Y$ by finitely many basic rectangles and form a tube] Fix $x \in X$. The slice $\{x\} \times Y$ is homeomorphic to $Y$ via $y \mapsto (x, y)$, hence compact. For each $y \in Y$, there exists $\alpha$ with $(x, y) \in U_\alpha \times V_\alpha$, so the collection $\{V_\alpha : x \in U_\alpha\}$ covers $Y$. By compactness of $Y$, finitely many suffice: there exist $\alpha_1, \ldots, \alpha_k$ (depending on $x$) with $Y = V_{\alpha_1} \cup \cdots \cup V_{\alpha_k}$ and $x \in U_{\alpha_j}$ for each $j$. Define $W_x = U_{\alpha_1} \cap \cdots \cap U_{\alpha_k}$. This is open in $X$ (finite intersection of [open sets](/page/Open%20Set)) and $x \in W_x$. For any $(x', y') \in W_x \times Y$: since $y' \in V_{\alpha_j}$ for some $j$, and $x' \in W_x \subseteq U_{\alpha_j}$, we have $(x', y') \in U_{\alpha_j} \times V_{\alpha_j}$. Therefore \begin{align*} W_x \times Y \subseteq (U_{\alpha_1} \times V_{\alpha_1}) \cup \cdots \cup (U_{\alpha_k} \times V_{\alpha_k}). \end{align*} [/step] [step:Extract a finite subcover of $X$ from the tubes and assemble the final subcover] The collection $\{W_x\}_{x \in X}$ is an open cover of $X$ (since $x \in W_x$ for every $x$). By compactness of $X$, there exist $x_1, \ldots, x_m$ with $X = W_{x_1} \cup \cdots \cup W_{x_m}$. Then \begin{align*} X \times Y = (W_{x_1} \times Y) \cup \cdots \cup (W_{x_m} \times Y). \end{align*} Each $W_{x_i} \times Y$ is covered by finitely many basic open rectangles from the refinement. The union of these $m$ finite collections is a finite subcollection of basic open sets covering $X \times Y$. Selecting the corresponding members of $\mathcal{U}$ gives a finite subcover of the original cover. [/step]