Prove a basis for the product topology on $\prod_{a \in \lambda}X_{a}$ is the collection of all sets of the form $\prod_{a \in \lambda}U_a$ where $U_a$ is open in $X_a$ for each $a$ and $U_a=X_a$ for all but finitely many $a$. Given that the product topology is defined as the topology generated by the subbasis $S=\bigcup\limits_{B \in \lambda}S_B$. Where $S_B=\{\pi_B^{-1}(U_B)|U_B \ \ \text{open in}\ \ X_B \}$
Attempt:Let $S=\bigcup\limits_{B \in Y}S_B$ be a subbasis for the product topology, where $S_B=
\{ \pi^{-1}_{B}(U_B)|U_{B} \ \ \text{open in}\ \ X_B \}$. Then each of the factors of elements of $S$ are of the form $U_a=X_a$ for all $a \neq B$ and $U_B$ for $B \in Y$.So only one factor is not necessarily equal to $X_a$. Then a basis element is a finite intersection of subbasis elements $\pi^{-1}_i(U_i)$ where $i=i_1,…,i_k$. So $U_a=X_a$ for all $i\neq i_1,…,i_k$.
Best Answer
Yes, that's essentially the idea: more succinctly:
(standard fact about bases and subbases): if $\mathcal{S}$ is a subbase, then the set of finite intersection of elements from $\mathcal{S}$ form a base.
A finite intersection from $\mathcal{S}$ looks like $\pi_{a_1}^{-1}[U_1] \cap \pi_{a_2}^{-1}[U_2] \ldots \cap \pi_{a_n}^{-1}[U_n]$, where $a_1, \ldots a_n$ form a finite subset of $\lambda$ (all distinct, because we can use $p_a^{-1}[U] \cap \pi_a^{-1}[V]= \pi_a^{-1}[U \cap V]$ to take sets with the same projection together, a minor point that you omitted) and this is just the set $$\prod_{a \in \lambda} O_a$$ where $O_a = U_i$ if $a = a_i$ for some $i \in \{1,\ldots n\}$ (the finite exception set) and $O_a = X_a$ otherwise. So these form exactly the base you had to prove to be a base: it's just the standard base from the subbase.