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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 $X^2$?

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This (rather exotic) covering property is often called "Property D". Here 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.

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