Algernon's argument seems to need a special case of Tychonoff's theorem to get to the finiteness of the union. Here is an argument which avoids that.

Lemma: Let $A\subseteq X$ be compact and $B\subseteq Y$ arbitrary. Let $W\subseteq X\times Y$ be open such that for each $x\in A$ there are a neighborhoods $U_x$ of $x$ and $V_x$ of $B$ with $U_x\times V_y\subseteq W$, then there are neighborhoods $U$ of $A$ and $V$ of $B$ with $U\times V\subseteq W$.

For the proof of the lemma, one can restrict by the compactness to finitely many neighborhoods and apply Algermon's argument with $V$ being a finite intersection.

To see that the hypotheses of the lemma are satisfied if $B$ is compact with $A\times B\subseteq W$, the lemma itself can be applied (with exchanged roles of $X$ and $Y$, $A:=B$ and $B:=\{x\}$): The hypothesis of the lemma is then satisfied by the mere definition of the product topology.

Algernon's argument seems to need a special case of Tychonoff's theorem to get to the finiteness of the union. Here is an argument which avoids that.

Lemma: Let $A\subseteq X$ be compact and $B\subseteq Y$ arbitrary. Let $W\subseteq X\times Y$ be open such that for each $x\in A$ there are a neighborhoods $U_x$ of $x$ and $V_x$ of $B$ with $U_x\times V_x\subseteq W$, then there are neighborhoods $U$ of $A$ and $V$ of $B$ with $U\times V\subseteq W$.

For the proof of the lemma, one can restrict by the compactness to finitely many neighborhoods and apply Algermon's argument with $V$ being a finite intersection.

To see that the hypotheses of the lemma are satisfied if $B$ is compact with $A\times B\subseteq W$, the lemma itself can be applied (with exchanged roles of $X$ and $Y$, $A:=B$ and $B:=\{x\}$): The hypothesis of the lemma is then satisfied by the mere definition of the product topology.