Suppose that you have a family of topological spaces $(X_i)_{i\in I}$, and for each $i$, a base $\mathcal{B}_i$ of $X_i$. Is it true that the set of all products $\prod_i A_i$ where each $A_i$ is in $\mathcal{B}_i$, and for all but possibly a finite number of $i$, $A_i=X_i$, is a base of the product space $\prod_i X_i$ ?
I've seen that statement in the case finite products, so I'm wondering if it's true in the general case.
Yes, if $O$ is open in $\prod_i X_i$ and $x \in O$, we have a standard basic open $\prod_i U_i$ with all $U_i = X_i$ except for a finite subset $F$ of $I$, where $U_i$ is open in $X_i$, such that $x \in \prod_i U_i \subseteq O$. This is by definition of the product topology.
Now for each $i \in F$ we have $x_i \in U_i$ so there is some $B_i$ in the base $\mathcal{B}_i$, such that $x_i \in B_i \subseteq U_i$, by the definition of a base.
Now $\prod_i B_i$ with $B_i$ as above for $i \in F$ and $B_i = X_i$ for $i \notin F$ is in your collection and obeys $x \in \prod_i B_i \subseteq \prod_i U_i \subseteq O$, and as $O$ and $x \in O$ were arbitary, this shows that collection is indeed a base.
In particular, if $I$ is countable and all $X_i$ are second countable, so is $\prod_i X_i$, using this base formed from countable bases for all $X_i$.