= Solution
The <product topology> on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. Equivalently, these rectangles form a <basis>.
The empty set is the empty union, and $X\times Y$ is itself an allowed rectangle. An arbitrary union of sets of this form is still a union of allowed rectangles. Finally,
$$
(U_1\times V_1)\cap(U_2\times V_2)=(U_1\cap U_2)\times(V_1\cap V_2),
$$
which is an allowed rectangle because the factor topologies are closed under finite intersections. Distributing intersection over unions shows that the intersection of any two product-open sets is product-open; induction handles all finite intersections. These verify every <topology axiom>. Rectangles alone are generally not a topology, which is why their arbitrary unions are included.
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