If each $(X_i,T_i)$ is finite and discrete, then the box product is not compact.
Let each $(X_i,T_i) = \{x_i\}$, then the box product is only covered by the open set $U = \prod \{x_i\}$, but since this is also the finite subcover for the box product, it is compact. This seems to be a contradiction to the first statement.
Could someone explain to me what I'm not understanding here?