 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,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.