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  "body": "\n## Proof\n\nWe use the open cover condition and the tube lemma argument.\n\n1. Let $(W_j)_{j \\in J}$ be open subsets of $X \\times Y$ with $K \\times L \\subseteq \\bigcup_{j \\in J} W_j$. By [[compactness via open covers|0P6PMJB4MA]], we must cover $K \\times L$ by finitely many $W_j$.\n2. If $K$ or $L$ is empty, then $K \\times L$ is empty, and the empty subfamily covers it. From now on let $K$ and $L$ be nonempty.\n3. Let $(x, y) \\in K \\times L$, and choose $j(x, y) \\in J$ with $(x, y) \\in W_{j(x, y)}$.\n4. The sets $U \\times V$ with $U$ open in $X$ and $V$ open in $Y$ form a [[basis|06EX0FHBTF]] of the [[product topology|0Z57EC6YX3]]. So there are open $U_{x, y} \\subseteq X$ and $V_{x, y} \\subseteq Y$ with $(x, y) \\in U_{x, y} \\times V_{x, y} \\subseteq W_{j(x, y)}$.\n5. Fix $x \\in K$. The open sets $V_{x, y}$, for $y \\in L$, cover $L$, since $y \\in V_{x, y}$.\n6. Since $L$ is compact, there are $y_1, \\ldots, y_m \\in L$ with $L \\subseteq V_{x, y_1} \\cup \\cdots \\cup V_{x, y_m}$. Here $m \\ge 1$, because $L$ is nonempty.\n7. Let $T_x = U_{x, y_1} \\cap \\cdots \\cap U_{x, y_m}$. It is open in $X$ by the axioms of a [[topology|0CNQXVBHCX]], and it contains $x$.\n8. Let $J_x = \\set{j(x, y_1), \\ldots, j(x, y_m)}$, a finite subset of $J$. Claim: $T_x \\times L \\subseteq \\bigcup_{j \\in J_x} W_j$. Let $(a, b) \\in T_x \\times L$. By step 6, $b \\in V_{x, y_k}$ for some $k$.\n9. Also $a \\in T_x \\subseteq U_{x, y_k}$. So $(a, b) \\in U_{x, y_k} \\times V_{x, y_k} \\subseteq W_{j(x, y_k)}$ by step 4. This proves the claim.\n10. The open sets $T_x$, for $x \\in K$, cover $K$. Since $K$ is compact, there are $x_1, \\ldots, x_p \\in K$ with $K \\subseteq T_{x_1} \\cup \\cdots \\cup T_{x_p}$.\n11. Let $(a, b) \\in K \\times L$. Then $a \\in T_{x_l}$ for some $l$, so $(a, b) \\in T_{x_l} \\times L$.\n12. By step 8, $(a, b) \\in W_j$ for some $j \\in J_{x_l}$.\n13. So the sets $W_j$ with $j$ in the finite set $J_{x_1} \\cup \\cdots \\cup J_{x_p}$ cover $K \\times L$. By step 1, $K \\times L$ is compact.\n\n## Idea\n\nFor a fixed $x$, the slice $\\set{x} \\times L$ is a copy of the compact set $L$, so finitely many basic boxes cover it. Their first factors meet in an open set $T_x$, which gives a whole tube $T_x \\times L$ covered by finitely many sets. Compactness of $K$ then needs only finitely many tubes.\n",
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