Box topology is rarely metrizable

Proof. We consider two cases.
Case 1: $X$ is Hausdorff. We will show that $\prod_i X$ is not first countable, hence not metrizable. Let $x$ be a non-isolated point in $X$ (which exists by the assumption that $X$ is not discrete). For the sake of contradiction assume that $\{\, A_n : n \in \N \,\}$ is coutable at $(x_i)_{i \in I}$ where $x_i = x$ for each $i \in I$. For any $A_n$ we find a basic open \nei which contains $(x_i)$, namely $$A_n \supset \prod_i U^{(n)}_i \ni (x_i),$$ where each $U^{(n)}_i$ is an open \nei of $x_i$. Since $I$ is infinite we take an infinite countable subset $I' \subset I$. For convinience relabel elements of $I$ so that $I' = \{i_1, i_2 \dots\}$.

For any $n \in \N$ let $y^{(n)} \in U_{i_n}^{(n)} \subset X$ be a point distinct from $x$. Such a point exists because we assumed that $x$ is not isolated. Since $X$ is Hausdorff we can separate $x$ from $y^{(n)}$ with an open \nei $V_n$ of $x$, so that $y^{(n)} \not \in V_n$. Let $$V = \prod_i \begin{cases} V_n, \quad &\text{for }i = i_{n},\\ X, \quad &\text{otherwise.} \end{cases}$$ Set $V$ is a basic open \nei of $(x_i)$. It is enough to show that $A_n \not \subset V$ for any $n \in \N$. Assume that for some natural number $k$ we have $A_k \subset V$. For $i \in I$ let $$y_i = \begin{cases} y^{(k)}, \quad &\text{for } i = i_k, \\ x, \quad &\text{otherwise.} \end{cases}$$ and define $(y_i) \in \prod_i X$. Clearly $(y_i) \in \prod_i U^{(k)}_i \subset A_k$. On the other hand $$y_{i_k} = y^{(k)} \not \in V_k$$ by construction of $V_k$. Hence $(y_i) \not \in V$. We arrive at a contradiction with the inclusion $A_k \subset V$.

Case 2. $X$ is not Hausdorff. It is enough to show that $\prod_i X$ with the box topology is not Hausdorff. Let $x,y$ be the points such that for any open \nei{s} $U, V$ of $x,y$ (respectively) we have $U \cap V \neq \emptyset$. Then points $(x_i), (y_i)$ where $x_i = x$ and $y_i = y$ cannot be separated by basic open sets in the box topology. ◻