Theorem 15.2 of Munkres Topology

You're making this way too hard. Let $\mathcal{T}$ be the topology generated by the subbase $\mathcal{S}$, and let $\mathcal{T}_p$ be the product topology on $X \times Y$.

All members of $\mathcal{S}$ are subsets of $X \times Y$ and in particular we can see that $X \times Y \in \mathcal{S}$ (we can choose $U=X$ in the first type, and also $V=Y$ in the second type). So that $\mathcal{S}$ is a subbase (in the Munkres sense) is totally obvious.

All members of $\mathcal{S}$, be they of the form $U \times Y$ or $X \times V$ are in $\mathcal{T}_p$ by definition. So trivially $\mathcal{T} \subseteq \mathcal{T}_p$. And if we look at the base $\mathcal{B}$ generated by $\mathcal{S}$, we get all sets $(U \times Y) \cap (X \times V) = U \times V$ in $\mathcal{B}$ so that the base generated by $\mathcal{S}$ is precisely the standard base for $\mathcal{T}_p$ so we have equality of these topologies. That's really all you have to realise.

Two topologies with the same base (basis if you prefer) are equal...