product-$\sigma$-algebra generated by ... Let $I$ be a countable Indexset, $(X_i,\mathcal{A}_i)\: (i \in I)$ be measurable spaces and $X:= \prod\limits_{i \in I}X_i$; $\:\mathcal{A}:=\bigotimes\limits_{i \in I} \mathcal{A}_i$ the product space and the product-$\sigma$-algebra.
Does the set $\{\prod\limits_{i \in I} A_i \mid A_i \in \mathcal{A_i}\}$ already generate the product-$\sigma$-algebra $\mathcal{A}$?
We have to show: $\sigma(\{A_k\times \prod\limits_{i \in I, i \neq k}X_i \mid A_i \in \mathcal{A_k}  \}) = \sigma(\{\prod\limits_{i \in I}A_i \mid A_i \in \mathcal{A}_i\} )   $.
"$\subseteq$": Let $x \in \{A_k\times \prod\limits_{i \in I, i \neq k}X_i \mid A_i \in \mathcal{A_k}\}$ then since $X_i \in \mathcal{A}_i$, there holds $x \in \{\prod\limits_{i \in I}A_i \mid A_i \in \mathcal{A}_i\}$ and hence $\sigma(\{A_k\times \prod\limits_{i \in I, i \neq k}X_i \mid A_i \in \mathcal{A_k}  \}) \subset \sigma(\{\prod\limits_{i \in I}A_i \mid A_i \in \mathcal{A}_i\} )$
"$\supseteq$": $ \{\prod\limits_{i \in I}A_i \mid A_i \in \mathcal{A}_i\}= \{\bigcap\limits_{i \in I}\pi_{i}^{-1}(A_i)\mid A_i \in \mathcal{A}_i\} \subset \mathcal{A} $ and hence $\sigma(\{A_k\times \prod\limits_{i \in I, i \neq k}X_i \mid A_i \in \mathcal{A_k}  \}) \supseteq \sigma(\{\prod\limits_{i \in I}A_i \mid A_i \in \mathcal{A}_i\} )   $.
Is this right?
 A: If the product $\sigma$-algebra on $X$ is defined as the smallest $\sigma$-algebra that contains all $\{\pi_i^{-1}[A]: A \in \mathcal{A}_i\}$ as you are using, then this argument is valid, I would say.
Note that the countability of $I$ is essentially used in the second inclusion.
A: In this answer I would like to make propaganda for the following definition of the product $\sigma$-algebra $\mathcal A=\bigotimes\limits_{i \in I} \mathcal{A}_i$ on a product space $X=\prod_{i\in I}X_i$: 
Further in my answer I also handle the case where $I$ is not countable.

$\mathcal A=\bigotimes\limits_{i \in I} \mathcal{A}_i$ it is the smallest $\sigma$-algebra on $X=\prod_{i\in I}X_i$ such that the projection $\pi_i:X\to X_i$ is measurable for every $i\in I$.

It is not difficult to verify that this definition leads to the same $\sigma$-algebra as the usual one.
Now I will give an alternative approach to your question with this definition in our luggage.

Theorem:
Let it be that $\mathcal C$ denotes the collection of sets of the form $\prod_{i\in I}A_i$ with $A_i\in\mathcal A$ for every $i\in I$ and let $\mathcal B:=\sigma(\mathcal C)$, i.e. the $\sigma$-algebra generated by $\mathcal C$.
Then $\mathcal A\subseteq\mathcal B$.
Under the presumption that the measurable spaces $(X_i,\mathcal A_i)$ are not trivial (i.e. a set $A_i\in\mathcal A_i$ exists with $A_i\notin\{\varnothing,X_i\}$) we will have $\mathcal A=\mathcal B$ if and only if $I$ is countable.

Proof: 
For proving that $\mathcal A\subseteq\mathcal B$ it is only needed to show that the projections $\pi_i:X\to X_i$ are all measurable if $X$ is equipped with $\sigma$-algebra $\mathcal B$. This is not difficult because for every $i\in I$ and every $U\in\mathcal A_i$ it is evident that $\pi_i^{-1}(U)\in\mathcal C\subseteq\mathcal B$.
Now let $I$ be countable and let $U\in\mathcal C$ so that sets $A_i\in\mathcal A$ exist with $U=\bigcap_{i\in I}\pi_i^{-1}(A_i)$. This shows that we can write $U$ as a countable intersection of elements of $\mathcal A$ which justifies the conclusion that $U\in\mathcal A$. So a countable $I$ guarantees that $\mathcal C\subseteq\mathcal A$ and consequently $\mathcal B\subseteq\mathcal A$. Combining this with what was proved above we find that $\mathcal A=\mathcal B$ if $I$ is countable.
Finally let it be that $I$ is not countable. For every countable $J\subset I$ we have the $\sigma$-algeba $\mathcal A_J$ characterized by the property that it is the smallest $\sigma$-algebra on $X$ such that all projections $\pi_i$ with $i\in J$ are measurable. Now let $\wp_{\text{co}}(I)$ denote the collection of all countable subsets of $I$ and have a look at the collection $\bigcup_{J\in\wp_{\text{co}}(I)}\mathcal A_J$. It is not difficult to prove that this collection is a $\sigma$-algebra and that for every $i\in I$ the projection $\pi_i$ is measurable wrt to this $\sigma$-algebra. Also it is clear that every $\sigma$-algebra with this property must contain this collection, so our conclusion is that $\mathcal A=\bigcup_{J\in\wp_{\text{co}}(I)}\mathcal A_J$. However if $A_i\in\mathcal A_i$ with $A_i\notin\{\varnothing,X_i\}$ for every $i\in I$ then the set $\prod_{i\in I}A_i$ is  an element of $\mathcal B$ but is not an element of $\mathcal A$. Proved is now that in that case $\mathcal A\neq\mathcal B$.
