Let $I$ be uncountable. Suppose each $X_i$ is Hausdorff with atleast two distinct points, $p_i,q_i$. Put $X=\prod_{i\in I}X_i$ (product topology). Then $X$ cannot be first countable.
Proof: Suppose for a contradiction that $X$ is first countable. Fix $f\in \prod_iX_i$. Suppose it has a nested local basis, $(U_n)_{n\in \mathbb{N}}$.
Claim: There exists an open set $U$ such that $f\in U$, but for all $n$, $U_n$ is not contained in $U$.
The proof I read was something like:
Step 1: Recall that the product topology has a basis. Observe that, for each $n$ there exists a basis element, $B_n$, such that $f\in B_n \subseteq U_n$. Note that $B_n$ is the finite intersection of sets of the form $\pi_i^{-1}(U_i)$ where $\pi_i$ denotes the projection map and $U_i$ is open in $X_i$. So, define $D$ to be the the set of indices in the finite intersection for each $n$. Therefore, $D$ is countable, being the countable union of finite sets.
Step 2: Therefore, we may choose some $P\in I$ not in $D$. Consider $X_P$. Since $X_P$ has two distinct points, we may suppose $q_P$ is different from $\pi_P(f)=f(P)$. Since $X_P$ is hausdorff, let $M_1$ be a neighborhood of $f(P)$ and $M_2$ a neighborhood of $q_P$, such that $M_1\cap M_2=\varnothing$. Put $U=\prod_{i\in I}O_i$ where $O_i= M_1$ if $i=P$ and $X_i$ if $i\neq P$. This set is open and contains $f$.
But... I can't see why $U_n$ is not contained in $U$?