[proofplan] We prove $X \times Y$ is compact via the tube lemma technique. Given an open cover of $X \times Y$, for each $x \in X$ we cover the compact slice $\{x\} \times Y$ by finitely many basic open rectangles, then form an open tube $W_x \times Y$ by intersecting the $X$-components. The tubes $\{W_x\}$ cover $X$; compactness of $X$ yields a finite subcover, from which we assemble a finite subcover of $X \times Y$. [/proofplan] [step:Cover each compact slice $\{x\} \times Y$ by finitely many basic open rectangles] Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $(x, y) \in X \times Y$, choose $U_{(x,y)} \in \mathcal{U}$ containing $(x, y)$. By definition of the product [topology](/page/Topology), there exist [open sets](/page/Open%20Set) $A_{(x,y)} \subseteq X$ and $B_{(x,y)} \subseteq Y$ with $(x, y) \in A_{(x,y)} \times B_{(x,y)} \subseteq U_{(x,y)}$. Fix $x \in X$. The map $y \mapsto (x, y)$ is a homeomorphism from $Y$ onto $\{x\} \times Y$, so $\{x\} \times Y$ is compact. The collection $\{A_{(x,y)} \times B_{(x,y)}\}_{y \in Y}$ covers $\{x\} \times Y$, so by compactness, finitely many suffice: there exist $y_1, \ldots, y_m \in Y$ with \begin{align*} \{x\} \times Y \subseteq \bigcup_{j=1}^{m} A_{(x,y_j)} \times B_{(x,y_j)}. \end{align*} [/step] [step:Form the tube $W_x \times Y$ by intersecting the $X$-components] Define $W_x = A_{(x,y_1)} \cap \cdots \cap A_{(x,y_m)}$. This is a finite intersection of open sets in $X$, hence open, and $x \in W_x$ (since $x \in A_{(x,y_j)}$ for every $j$). We verify $W_x \times Y \subseteq \bigcup_{j=1}^{m} U_{(x,y_j)}$. Let $(x', y') \in W_x \times Y$. Since $\{B_{(x,y_j)}\}_{j=1}^{m}$ covers $Y$ (the finite subcover from the previous step), there exists $j$ with $y' \in B_{(x,y_j)}$. Since $x' \in W_x \subseteq A_{(x,y_j)}$, we have $(x', y') \in A_{(x,y_j)} \times B_{(x,y_j)} \subseteq U_{(x,y_j)}$. [/step] [step:Extract a finite subcover of $X \times Y$ using compactness of $X$] The collection $\{W_x\}_{x \in X}$ is an open cover of $X$. By compactness of $X$, there exist $x_1, \ldots, x_k$ with $X = W_{x_1} \cup \cdots \cup W_{x_k}$. Then \begin{align*} X \times Y = \bigcup_{i=1}^{k} (W_{x_i} \times Y) \subseteq \bigcup_{i=1}^{k} \bigcup_{j=1}^{m_i} U_{(x_i, y_j)}. \end{align*} Each tube contributes finitely many sets from $\mathcal{U}$, and there are finitely many tubes. The union is a finite subcollection of $\mathcal{U}$ covering $X \times Y$. [guided] Let us trace the structure of the argument. Compactness of $X \times Y$ is proved by applying compactness of $Y$ and compactness of $X$ in sequence, each time reducing an infinite collection to a finite one. **First reduction (compactness of $Y$):** For a fixed $x \in X$, the slice $\{x\} \times Y$ is homeomorphic to $Y$ (hence compact) and is covered by basic open rectangles $A_{(x,y)} \times B_{(x,y)}$ from the cover. Compactness of $Y$ selects finitely many: $y_1, \ldots, y_m$ with $Y = B_{(x,y_1)} \cup \cdots \cup B_{(x,y_m)}$. **Tube construction:** The finite intersection $W_x = \bigcap_{j=1}^m A_{(x,y_j)}$ "fattens" the slice into a tube: for any $(x', y') \in W_x \times Y$, some $j$ has $y' \in B_{(x,y_j)}$, and $x' \in W_x \subseteq A_{(x,y_j)}$, so $(x', y') \in A_{(x,y_j)} \times B_{(x,y_j)} \subseteq U_{(x,y_j)}$. The tube $W_x \times Y$ is covered by finitely many members of $\mathcal{U}$. **Second reduction (compactness of $X$):** The tubes $\{W_x\}_{x \in X}$ form an open cover of $X$. Compactness of $X$ selects finitely many: $x_1, \ldots, x_k$. Each tube carries finitely many cover elements, so the total is finite. The key insight is that the intersection $W_x = \bigcap_j A_{(x,y_j)}$ must be finite to guarantee openness. An infinite intersection of open sets need not be open, which is why compactness of $Y$ (reducing to finitely many $y_j$) is essential. [/guided] [/step]