Showing a collection generates a sigma algebra Let $X = X_1 \times X_2 $, let $\pi_i : X \to X_i $ ($i=1,2$) be coordinate maps. Let $\mathcal{M}_i $ be $\sigma-algebra $ on $X_i$. Let $\mathcal{F} = \{ \pi_i^{-1}(E_i) : E_i \in \mathcal{M}_i \} $. $\sigma( \mathcal{F} ) $ is  $\sigma-$algebra on the product $X$.
PROBLEM: Show that $\sigma( \mathcal{F} ) $ is also generated by $ \mathcal{E} = \{ E_1 \times E_2 : E_i \in \mathcal{M}_i \} $
attempt:
Lets take an element from $\mathcal{F} $(which also belongs to $\sigma( \mathcal{F} ) $), say $\pi_1^{-1}(E_1) $. We know
$$ \pi_1^{-1}(E_1) = \{ (x_1,x_2) : x_1 \in E_1, \; \; x_2 \in X_2 \}$$
If we have that $E_2 = X_2$, then it must be the case that $\pi_1^{-1}(E_1) = E_1 \times E_2 $, and so $\pi_1^{-1}(E_1) \in \mathcal{E}  $. Hence, $\sigma( \mathcal{F} ) \subset \mathcal{E} $. But we know $\mathcal{E} \subset \sigma( \mathcal{E} ) $. In particular, we have $\boxed{ \sigma(\mathcal{F}) \subset \sigma(\mathcal{E}) } $
For the other direction, we take one element from $\mathcal{E} $, and notice
$$E_1 \times E_2 = \{ (x_1,x_2) : x_i \in E_i \} = \{ (x_1,x_2) : x_1 \in E_1, x_2 \in X_2 \} \cap \{ (x_1,x_2) : x_1 \in X_1 , x_2 \in E_2 \} = \pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \mathcal{F} \subset \sigma( \mathcal{F} ) $$
and so we obtain that $\mathcal{E} \subset \sigma( \mathcal{F} )$ which implies that $\sigma( \mathcal{E} ) \subset \sigma( \mathcal{F} ) $ which in turn gives us that
$$ \sigma( \mathcal{E} ) = \sigma( \mathcal{F} ) $$
as was to be shown.
Is this a correct proof ? Can we generalize this to any indexed collection?
 A: This can only be generalized to a countable indexed collection, by the result of your earlier question. The first part is ok. About the second part, it is not true that $\pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \mathcal{F}$, in general. 
I'll just copy what we did there, in the notation here (and do notice how similar things are).
We want to prove that $\sigma({\cal F}) \subset \sigma({\cal E})$: take $E_1 \in {\cal M}_1$ and $E_2 \in {\cal M}_2$. Then $$\pi_1^{-1}(E_1) = E_1 \times X_2 \in {\cal E}, \quad \pi_2^{-1}(E_2) = X_1 \times E_2 \in {\cal E}.$$Since my $E_i$ were arbitrary, we get: $${\cal F} \subseteq {\cal E}\subseteq \sigma({\cal E}) \implies \sigma({\cal F})\subseteq \sigma({\cal E}).$$On the other hand, take $E_1 \times E_2 \in {\cal E}$, arbitrary: we have: $$E_1 \times E_2 = \pi_1^{-1}(E_1) \cap \pi_2^{-1}(E_2) \in \sigma({\cal F}),$$because the RHS is a countable (two!) intersection of measurables. So: $${\cal E} \subseteq \sigma({\cal F}) \implies \sigma({\cal E})\subseteq \sigma({\cal F}),$$and we conclude that $\sigma({\cal E})=\sigma({\cal F})$.
