$(X,T_X)$ and $(Y,T_Y)$ are topological spaces. Let $Z=X\times Y$ and $$B_Z=\{ U\times V: U\subseteq X\ \text{open and}\ V\subseteq Y \ \text{open}\}$$.
Prove that $B_Z$ is a base for some topology $T$ on $Z$, where $T$ consists of all unions $\displaystyle{\bigcup_{\alpha\in A}W_\alpha} $ of sets $W_\alpha\in B_Z$.
We want to show every open subset is a union of a member of $B_Z$.
So, we need to show $\forall\ \text{open}\ S\subseteq Z, \exists\ U\subseteq X\ , V\subseteq Y$ ($U$ and $V$ are open) such that $S=U\times V$. Is this correct? I have no idea how to prove this.
A collection $\mathcal{B}$ of subsets of a set $X$ is a base for a topology on $X$ iff
In that case the set of all unions of members of $\mathcal{B}$ is exactly the topology that $\mathcal{B}$ is a base of (and the minimal topology that contains all members of $\mathcal{B}$).
In your case, the whole space is $Z$ and $Z= X \times Y \in \mathcal{B}_Z$ is already in the base, so Nr 1. is trivial.
Nr 2. is easy too: $B_1 \cap B_2$ is of the form $(U \times V) \cap (U' \times V') = (U \cap U') \times (V \cap V')$, so if we have $x$ in the intersection on the left we can just take the same intersection as $B_3$ which is in $\mathcal{B}_Z$ by the equality and the fact that topologies are closed under finite intersections.
Then by definition a set in the product topology (i.e. the topology generated by $\mathcal{B}_Z$) is a union of sets of the form $U \times V$, it will not always be of that exact form (open circles in the plane e.g. or the plane minus the origin) and this is not what was asked by the question. This was just to show $\mathcal{B}_Z$ is a base for a topology and this comes down to checking the two standard conditions.