Cofinite topology on $X \times X$ with $X$ an infinite set Let $X$ be an infinite set. And consider $(X \times X)_{cof}$ and $X_{cof} \times X_{cof}$. I can see that $(X \times X)_{cof}$ is not finer than $X_{cof} \times X_{cof}$. 
MY QUESTION: But is $X_{cof} \times X_{cof}$ finer (hence strictly finer) than $(X \times X)_{cof}$ ?
Specifically, an open set $U$ in $(X \times X)_{cof}$ will be such that $(X\times X)-U$ is finite, meaning here that $(X\times X)-U$ is some finite set of ordered pairs. But I'm struggling to formally show that such a set can be written as an open set in $X_{cof} \times X_{cof}$, i.e., as a union of products of two sets whose complements are finite subsets of $X$. I'd appreciate any help.

Maybe is slightly simpler to think about closed sets. For any pair of points $x,y\in X$, both $\{x\}$ and $\{y\}$ are closed in $X_{cof}$. By definition of product topology, both $\{x\}\times X$ and $X\times \{y\}$ are closed in $X_{cof}\times X_{cof}$. Therefore their intersection
$$(\{x\}\times X)\cap (X\times \{y\})=\{(x,y)\} $$
is closed. In particular, since finite unions of closed sets are closed, every finite set in $X\times X$ is closed in $X_{cof}\times X_{cof}$. This proves that $X_{cof}\times X_{cof}$ is finer than $(X\times X)_{cof}$.
It is also strictly finer since for instance $\{x\}\times X$ is closed in $X_{cof}\times X_{cof}$, but since it is infinite it can not be closed in $(X\times X)_{cof}$.
