Caution: Answer contains a serious mistake (see comments below by Joseph van Name), but is not deleted for the record.
A space $(X,\tau)$ is called a $D$-space if if whenever one is given a neighborhood $N(x)$ of $x$ for each $x\in X$, then there is a closed discrete subset $D\subseteq X$ such that $\{N(x): x\in D\}$ covers $X$. This is a survey of $D$-spaces and their properties.
A routine verification shows that whenever $X_i$ are $D$-spaces for all $i$, then so is $\prod_{i\in I}^{\textrm{Box}}X_{i}$.
EDIT: Verification of the claim above.
Let $X_i, i\in I$ be $D$-spaces and assume $\{N\big((x_i)\big):(x_i) \in\prod_{i\in I} X_i\}$ is an assignment of a neighborhood to each point of $\prod_{i\in I} X_i$ with the box product topology. We can assume that $N\big((x_i)\big) = \prod_{i\in I} N_i(x_i)$ for each $i$ where $N_i(x_i)$ is an open neighborhood of $x_i\in X_i$. Since each $X_i$ is a $D$-space let $D_i$ be a discrete set in $X_i$ such that $\bigcup \{N_i(d): d\in D_i\} = X_i$, for each $i$. So we set $$D = \prod_{i\in I} D_i.$$
Let $(y_i)\in \prod_{i\in I} X_i$. For each $i\in I$ we have $y_i \in N_i(d_i)$ for some $d_i\in X_i$ because by assumption $\bigcup \{N_i(d): d\in D_i\} = X_i$. So let $d:= (d_i)$, so $d\in D$. Then $(y_i) \in N\big((d_i)\big)$. This implies that $$\bigcup \{N\big((x_i)\big): (x_i)\in D\} = \prod_{i\in I} X_i.$$ Therefore $\prod_{i\in I} X_i$ with the box product topology is a $D$-space.