Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces,
$\mathcal{P}$ be a $\pi$-system on $X$ that generates $\mathcal{A}$, and $\mathcal{Q}$ be a $\pi$-system on $Y$ that generates $\mathcal{B}$. Suppose in addition that $X\in \mathcal{P}$ and $Y\in \mathcal{Q}$. Show that then $\mathcal{S} := \{ P\times Q: P\in \mathcal{P}, \, Q\in \mathcal{Q}\}$ is a $\pi$-system on $X\times Y$ that generates the $\sigma$-algebra $\mathcal{A}\times \mathcal{B}$, the $\sigma$-algebra generated by the algebra of finite disjoint unions of rectangles.
Please note that this question looks similar but is different from another post about $\sigma$-algebra shown on this link.
Using Dykin's $\pi-\lambda$ theorem, I managed to prove that $\sigma(\mathcal S) \subseteq \mathcal A \times \mathcal B$. However, I am having difficulty in trying to prove the reverse inclusion, namely $\mathcal A \times \mathcal B \subseteq \sigma(\mathcal S)$. It suffices to prove that $A \times B \in \sigma(\mathcal S)$ for any $A \in \mathcal A$ and $B \in \mathcal B$. Rather than using a complicated use of transfinite induction on the way that $A \in \mathcal A = \sigma(\mathcal P)$ and $B \in \mathcal B = \sigma(\mathcal Q)$ is built, is there a way to see that $A \times B \subseteq \sigma(\mathcal S)$?
This is similar to the proof of $\pi-\lambda$ theorem.
Consider $\{A\in \mathcal A: A\times B \in \sigma(S)\}$ where $B \in \mathcal Q$ is fixed. This is a $\lambda-$ system which contains $\mathcal P$. Hence it contains $\mathcal A$.
Now fix $A \in \mathcal A$ and consider $\{B\in \mathcal B: A\times B \in \sigma(S)\}$. This is a $\lambda-$ system containing $\mathcal Q$ and hence it contains $\mathcal B$. We are done.