= Solution
Use the <basis of a topology> consisting of open rectangles:
$$
\boxed{\mathcal B=\{U\times V:U\in\mathcal T_X,\ V\in\mathcal T_Y\}.}
$$
It covers $X\times Y$, since $X\times Y\in\mathcal B$. Whenever a point lies in two basis rectangles, their <intersection> is another basis rectangle containing it:
$$
(U_1\times V_1)\cap(U_2\times V_2)=(U_1\cap U_2)\times(V_1\cap V_2).
$$
These are precisely the two <basis of a topology> axioms. Arbitrary <unions> of the rectangles therefore form a <topology>, the <product topology>. To check the finite-intersection axiom directly, distribute the intersection of two unions over all pairs of rectangles and use the displayed formula. The empty union gives the <empty set>.
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