Generalization of the Tube Lemma (3.26.9)

James Munkres Topology, chapter 3.26, exercise 9:

Let $A⊆X$ and $B⊆Y$ be compact. Then show that for every open neighborhood $N⊆X×Y$ of $A×B$ we have $$A×B⊆U×V⊆N$$ for some open $U⊆X$ and $V⊆Y$.

Proof: We first show the partially generalized lemma when $A=\{a\}$ is a single point. Since $a×B≅B$ is compact, we obtain $$a×B⊆\bigcup(U_i×V_i)$$ when $U_i⊆X$ and $V_i⊆Y$ are open, for finitely many indices $i$. We may assume each $U_i$ is a neighborhood of $a$. We let $U=∩U_i$ and $V=∪V_i$ and claim they satisfy the lemma presented; this is evident since $U×V_i⊆U_i×V_i⊆N$ for all $i$ hence $U×V=∪(U×V_i)⊆N$, and also $a×b∈a×B$ implies $b∈V_i$ for some $i$ so $a×B⊆U×V$.

So by this partially generalized tube lemma, for each $a∈A$ let $U_a⊆X$ be a neighborhood of $a$ and $V_a⊆Y$ be open such that $U_a×V_a⊆N$. Since $A$ is compact, we may choose a finite subset $\mathcal{A}⊆A$ such that $A⊆∪_{α∈\mathcal{A}}U_α$. We let $U=∪U_α$ and $V=∩V_α$ and claim they satisfy the lemma presented. Since $U_α×V⊆U_α×V_α⊆N$ we see $U×V=∪(U_α×V)⊆N$. As well, for each $α∈\mathcal{A}$ we have $α×B⊆U_α×V_α$ so that $B⊆V_α$, implying $B⊆V$. And since the $U_α$ cover $A$ by construction we have $A×B⊆∪(U_α×V)⊆U×V$.$~\square$