Generalized Tube Lemma Let $X$ and $Y$ be compact Hausdorff spaces. And let $A\subset X$ and $B\subset Y$ be closed subspaces. If $U$ is open in $X \times Y$ and $U$ contains $A \times B$, then there exists open subsets $V \subset X$ and $W \subset Y$ such that $A \times B \subset V \times W \subset U$. 
I think this is an easier version of the generalized tube lemma since it assumes $X$ and $Y$ being Hausdorff, but I just can't seem to find such $V$ and $W$. I can find some open set in $X \times Y$ containing $A \times B$ but not able to show it is contained in $U$.
 A: (this is called the Wallace theorem/lemma in Engelking IIRC; there is a version for infinite products too, and Hausdorffness is not needed).
For each $a \in A$ and each $b \in B$ pick $V(a,b) \subseteq X$ open and $W(a,b) \subseteq Y$ open such that $$ (a,b) \in V(a,b) \times W(a,b) \subseteq U$$ which can be done as open squares form a base for the product topology. Now we go i two steps, exploiting both sets' compactness in turn.
Fix $b \in B$ for now, and note that $V(a,b)$, $a \in A$ forms an open cover of the compact $A$, so there are finitely many $F_b \subseteq A$ such that $V(b)=\cup_{a \in F_b} V(a,b)$ covers $A$ while $W(b) = \bigcap_{a \in F_b} V(a,b)$ is a neighbourhood of $b$. This defines open sets $V(b)\supseteq A$ of $X$ and open neighbourhoods $W(b)$ of $b$ such that $$V(b)\times W(b) \subseteq U$$. 
Now the $W(b)$  cover $B$, also compact, so finitely many, say $\{W(b): b \in H\}$ cover $B$, where $H \subseteq B$ is finite. Now define $V = \bigcap_{b \in H} V(b)$ and $W=\bigcup_{b \in H} W(b)$ and note that
$$A \times B \subseteq V \times W \subseteq U$$ as required.
It's just applying the idea of the tube lemma twice, really.
The proof only needs $A$ and $B$ to be compact, and nothing on $X$ and $Y$. So your textbook misses an easy generalisation here.
A: $X$ is a normal space, and $A\subseteq X\cap U$, which is open in $X$ so there is a $V$ open in $X$ such that $A\subseteq V\subseteq \overline V\subseteq  X\cap U.$ Similarly, there is a $W$ open in $Y$ such that $B\subseteq W\subseteq \overline W\subseteq  Y\cap U.$ Then, $A\times B\subseteq V\times W\subseteq (X\cap U)\times (Y\cap U)=U.$
