MY ATTEMPT
\begin{align*} (X\times Y)\cup (Z\times Y) & = \{(x,y)\mid x\in X\wedge y\in Y\}\cup\{(z,y)\mid z\in Z\wedge y\in Y\}\\\\ & = \{(w,y)\mid (w\in X\wedge y\in Y)\vee(w\in Z\wedge y\in Y)\}\\\\ & = \{(w,y)\mid (w\in X\vee w\in Z)\wedge y \in Y\}\\\\ & = \{(w,y)\mid (w\in X\cup Z)\wedge y \in Y\} = (X\cup Z)\times Y \end{align*}
Intuitively speaking, one can think of $X$, $Y$ and $Z$ as line segments contained in the real line. Thus the union of the rectangles $X\times Y$ and $Z\times Y$ is the rectangle $(X\cup Z)\times Y$. Any other way to solve it would be greatly appreciated. Thanks in advance.
Your proof looks correct. Instead of such an explicit, direct proof, I would have approached this more lazily. I would have taken an element of $(X \times Y)\cup(Z \times Y)$ and shown that it was in $(X\cup Z)\times Y$. Then taken an element in $(X\cup Z)\times Y$ and shown that it was in $(X \times Y)\cup(Z \times Y)$.