Why do we need set $A$ to be countable in following proposition? To define our terms: Let $\{ X_{\alpha} \}_{\alpha \in A } $ be any collection and $X = \prod_{\alpha \in A} X_{\alpha} $. Let $\pi_{\alpha}: X \to X_{\alpha} $ be coordinate maps and let $\mathcal{M}_{\alpha} $ be sigma algebras on $X_{\alpha} $. The product sigma algebra on $X$ is generated then by the  $\{ \pi_{\alpha}^{-1}(E_{\alpha} ) : E_{\alpha} \in \mathcal{M}_{\alpha}, \alpha \in A\}$, and we denote it by $\bigotimes_{\alpha \in A} \mathcal{M}_{\alpha} $.
Proposition:
If $A$ is $\textbf{countable}$, then $\bigotimes_{\alpha \in A} \mathcal{M}_{\alpha} $ is the sigma algebra generated by $\{ \prod_{\alpha \in A} E_{\alpha} : E_{\alpha} \in \mathcal{M}_{\alpha} \} $
Question:
Why do we need $A$ to be countable?? Isnt the proof works just fine if we work over any indexed collection?
 A: Let's analyze carefully the proof. By notation, and etc, I take it that you're using Folland's book to study. Let $\cal M$ be the $\sigma$-algebra generated by the products. He says:

If $E_\alpha \in {\cal M}_\alpha$, then $\pi_\alpha^{-1}(E_\alpha) = \prod_{\beta \in A}E_\beta$, where $E_\beta = X_\beta$ for $\beta \neq \alpha$;

This line gives: $$\{\pi_\alpha^{-1}(E_\alpha) \mid E_\alpha \in {\cal M}_\alpha,~\alpha \in A\}\subseteq \left\{\prod_{\alpha \in A}E_\alpha \mid E_\alpha \in {\cal M}_\alpha,~\alpha \in A\right\} \implies \bigotimes_{\alpha \in A}{\cal M}_\alpha \subseteq {\cal M}.$$Then he proceeds:

on the other hand, $\prod_{\alpha \in A}E_\alpha = \bigcap_{\alpha \in A}\pi_\alpha^{-1}(E_\alpha)$. The result therefore follows from Lemma $1.1$.

We now want to say: $$\left\{\prod_{\alpha \in A}E_\alpha \mid E_\alpha \in {\cal M}_\alpha\right\} \color{red}{\subseteq} \bigotimes_{\alpha \in A} {\cal M}_\alpha\implies {\cal M}\subseteq \bigotimes_{\alpha \in A}{\cal M}_\alpha,$$to finally conclude that: $${\cal M} = \bigotimes_{\alpha \in A}{\cal M}_\alpha.$$ The issue is that $\color{red}{\subseteq}$ above is only true if $A$ is countable.
