Question:
Let $\Omega_1$ and $\Omega_2$ be $2$ nonempty sets and let $\mathcal{A}$ be a $\sigma$-algebra of subsets of the Cartesian product $\Omega_1 \times \Omega_2$. Suppose that $\mathcal{S}_i$ is a collection of subsets of $\Omega_i$, $i = 1,2$. Assume that $\Omega_i \in \mathcal{S}_i$, $i = 1,2$. If $A \times B \in \mathcal{A}$ for all $A \in \mathcal{S}_1$ and all $B \in \mathcal{S}_2$, show that $A \times B \in \mathcal{A}$ for all $A \in \sigma(\mathcal{S}_1)$ and all $B \in \sigma(\mathcal{S}_2)$.
My attempt at the question:
We shall use fact that if $\Sigma_1$ is a $\sigma$-algebra on $\Omega_1$ such that $\mathcal{S}_1 \subseteq \Sigma_1$, then $\sigma(\mathcal{S}_1) \subseteq \Sigma_1$, and that if $\Sigma_2$ is a $\sigma$-algebra on $\Omega_2$ such that $\mathcal{S}_2 \subseteq \Sigma_2$, then $\sigma(\mathcal{S}_2) \subseteq \Sigma_2$.
So since $A \in \mathcal{S}_1 \subseteq \Sigma_1$ and $\sigma(\mathcal{S}_1)$ is the intersection of all $\sigma$-algebras on $\Omega_1$ containing $\mathcal{S}_1$, then $A \in \sigma(\mathcal{S}_1)$.
Likewise, since $B \ \in \mathcal{S}_2 \subseteq \Sigma_2$, then $B \in \sigma(\mathcal{S}_2)$.
Hence the statement that if $A \times B \in \mathcal{A}$ for all $A \in \mathcal{S}_1$ and all $B \in \mathcal{S}_2$, then $A \times B \in \mathcal{A}$ for all $A \in \sigma(\mathcal{S}_1)$ and all $B \in \sigma(\mathcal{S}_2)$ holds.
Problem:
I was wondering whether it is necessary to prove that $A \times B \in \mathcal{A}$ under the different conditions stated in the then-clause, since $A \times B \in \mathcal{A}$ has already been stated under the if-clause in the question. Also, have I missed out some crucial steps in the proof? And why is the assumption that $\Omega_i \in \mathcal{S}_i$, $i = 1,2$ necessary?
In your proof attempt, you have it backwards: Essentially, you start with $A \in {\cal S}_1$ and from this you conclude that $A \in \sigma({\cal S}_1)$ and so on. But you must start with an arbitrary $A \in \sigma({\cal S}_1)$ which in general will not be an element of ${\cal S}_1$.
Here's my idea:
Let ${\cal A}_1=\{A\subseteq\Omega_1\mid A\times \Omega_2\in{\cal A}\}$.
Clearly $\Omega_1\in{\cal A}_1$. If $A \in {\cal A}_1$, then $A \times \Omega_2\in{\cal A}$ and as $\cal A$ is a $\sigma$-algebra, we have $(\Omega_1 \setminus A) \times \Omega_2 = (\Omega_1 \times \Omega_2) \setminus (A \times \Omega_2)\in{\cal A}$ which implies $\Omega_1 \setminus A \in {\cal A}_1$. Finally, if $A_n\in{\cal A}_1$ for $n\in\mathbb N$, then $A_n\times \Omega_2\in{\cal A}$ for all $n$. Again, as $\cal A$ is a $\sigma$-algebra $(\bigcup_{n \in \mathbb N}A_n)\times \Omega_2=\bigcup_{n \in \mathbb N}(A_n\times \Omega_2)\in{\cal A}$ and thus $\bigcup_{n \in \mathbb N}A_n \in {\cal A}_1$. We have shown, that ${\cal A}_1$ is a $\sigma$-algebra.
Now, if $A \in {\cal S}_1$, then $A \times \Omega_2 \in \cal A$ by our assumption, so $A \in {\cal A}_1$. We've shown ${\cal S}_1\subseteq {\cal A}_1$ and therefore $\sigma({\cal S}_1)\subseteq{\cal A}_1$. (This is BTW why we need $\Omega_i\in{\cal S}_i$.)
Likewise, $\sigma({\cal S}_2)\subseteq{\cal A}_2$ with ${\cal A}_2=\{B\subseteq\Omega_2\mid \Omega_1\times B\in{\cal A}\}$.
Now, let $A\in\sigma({\cal S}_1)$ and $B\in\sigma({\cal S}_2)$. Then, by what we've just shown, $A\in{\cal A}_1$ and $B\in{\cal A}_2$, i.e. by definition $A \times \Omega_2 \in \cal A$ and $\Omega_1 \times B \in \cal A$. The intersection of these two sets is also in $\cal A$, and the intersection is $A \times B$.