We say that a space $X$ has *covering property (C)* if the following holds:


(C) For any open cover ${\cal U}$ of $X$ there is a closed discrete set $D\subseteq X$ and a map $\varphi: D\to {\cal U}$ such that


* $d\in\varphi(d)$ for all $d\in D$;
* $\bigcup \varphi(D) = X$.


If a space $X$ has property $(C)$, how about $X^2$?



If $X$ has the "discrete" covering property, how about $X^2$?
This (rather exotic) covering property is often called "Property D". [Here](http://auburn.edu/~gruengf/papers/dsurv7.pdf) you find a nice survey of $D$-spaces.


On p. 11, assuming $\textrm{(CH)}$, a space $Y$ is mentioned that has property $D$, but $Y^2$ doesn't.


So in ${\sf ZFC}$ you definitely won't be able to prove that whenever $X$ is a $D$-space, then so is $X^2$. However, I don't know whether an explicit counterexample exists in ${\sf ZFC}$, without assuming the Continuum Hypothesis.


