Hypotheses: Take infinitely many topological spaces $(X_i, \mathcal T_i)_{i\in I}$ such that none of $X_i$ is empty, and infinitely many $\mathcal T_i$ are strictly finer than than the indiscrete topology on $X_i$.
Claim: There exists an open set it the box topology $\mathcal T_\square$ that is not open in the product topology $\mathcal T_\Pi$.
Arguments: We first observe the kind of open sets that $\mathcal T_\Pi$ admits. It contains arbitrary unions of finite (nonepmty) intersections, each of which is of the form
$$
\bigcap_{i\in K} \biggl(\ \bigcap_{k = 1}^{N_i} \pi_i^{-1}(U_{i_k}) \biggr)\text,
$$
where $K$ is a finite nonempty subset of $I$ and $U_{k_i}\in\mathcal T_i$. (Also, $N_i\ge 1$ for each $i$.) Now, for each $i$, since $U_{i_k}$'s belong to the same $\mathcal T_i$, we can pull $\pi_i^{-1}$'s, getting
$$
\bigcap_{i\in K}\Biggl(\pi_i^{-1}\biggl( \underbrace{\bigcap_{k = 1}^{N_i} U_{i_k}}_{V_i} \biggr)\Biggr)\text.
$$
Since each $V_i$ is a finite nonempty intersections of open sets in $\mathcal T_i$, we have $V_i\in\mathcal T_i$. Hence we have gotten a simpler form $\bigcap_{i\in K}\Bigl(\pi_i^{-1}(V_i)\Bigr)$. It is not difficult to see that for any $W\subseteq X_i$, we have that $\pi_{i_0}^{-1}(W) = \prod_{i\in I} S_i$ where $S_i = X_i$ for $i\ne i_0$ and $S_{i_0} = W$. Using this and that the intersection and Cartesian products commute, we get the above as
$$
\prod_{i\in I} Y_i\text,
$$
where $Y_i = V_i$ if $i\in K$, otherwise $Y_i = X_i$. Notice that this Cartesian product has $Y_i\ne X_i$ in only finitely many $i$'s.
We are now to analyze an arbitrary union $\mathcal U$ of the above sets. Note that we will have only finitely many $i$'s such that the $i$-th coordinate of $\mathcal U$ will not range through all of $X_i$. (If we get an $X_i$ is the $i$-th coordinate in one of the intersections (of which $\mathcal U$ is a union of ), it'll "swamp" everything.) That is, $\pi_i(\mathcal U)\ne X_i$ for only finitely many $i$'s.
Now we just need to find a $\mathcal V\in\mathcal T_\square$ such that $\pi_i(\mathcal V)\ne X_i$ for infinitely many $i$'s, and we will be done. Just choose nenempty $U_i$'s from $\mathcal T_i$'s such that $U_i\ne X_i$ for infinitely many $i$'s and this can be done because of our hypotheses!