If $\tau$ is a topology different from the product topology, can the product be compact? Suppose $\tau$ is a topology on $\Pi_{i\in I}X_i$ where each $X_i$ is a topological space, different from the product topology such that the projection maps $\pi_i$ are continuous. 
I am trying to show that in this case, the product cannot be compact. I know that $\tau$ will contain the product topology, but I am not sure where to go from there?
 A: If the $X_i$ are Hausdorff then this holds, as $\prod_{i \in I} X_i$ under $\tau$ will be Hausdorff (being finer than the Hausdorff product topology) and thus not compact, because a finer topology than a Hausdorff compact one (i.e. the product topology) cannot be compact.
Compactness of the $X_i$ is necessary as all $X_i$ are continuous images of the product.
So within Hausdorff spaces the fuller argument goes: suppose $X$ (the product) is compact under $\tau$. Then all $X_i$ are compact being continuous images of $(X, \tau)$. It follows that $(X, \tau_p)$ (the product in the product topology) is Hausdorff (and compact). Then $i(x)=x$, $i:(X,\tau) \to (X,\tau_p)$ is continuous (as $\tau_p \subseteq \tau$), a bijection (obvious) and closed as a continuous map from a compact space to a Hausdorff one always is. So $i$ is a homoeomorphism, hence open, which implies $\tau \subseteq \tau_p$ and this contradicts our assumption that $\tau_p \subsetneq \tau$. So $(X, \tau)$ is not compact when we work in the class of Hausdorff spaces.
Maybe there is a topology on compact $T_1$ spaces $X_i$ such that $\tau$ finer than $\tau_p$ exists which is compact? That would be the best you could do..
A: Here's a simple counterexample in the case when the $X_i$ are not all Hausdorff. 
Let $A = \{0,1\}$ with the indiscrete topology $\{\emptyset, \{0,1\}\}$, and let $B$ be the one-point space $\{*\}$. Then the product topology on $A\times B$ is indiscrete ($A\times B$ is homeomorphic to $A$). If we instead equip $A\times B$ with the discrete topology, the projection maps $\pi_A$ and $\pi_B$ are still continuous (the discrete topology is finer than the product topology), and the space is still compact (since it is finite). 
