Past exam of the options I course of the University of Oxford / 2013 / E / 1 / b / ii / Solution

Choose $v_0\in V$, using its nonemptiness. For each $u\in U$, the openness of $U\times V$ in the product topology supplies open $O_u\subseteq X$, $W_u\subseteq Y$ such thatSince $v_0\in W_u$, every point of $O_u$ belongs to $U$. Thus $U={\bigcup }_{u\in U}{O}_u$ is open. Interchanging the factors, using a fixed $u_0\in U$, proves $V$ open. Therefore $\boxed{U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y}$. Nonemptiness matters: an empty product is open whatever the other factor is.