A ZFC example of a Menger space which is not Scheepers

Question

$\Omega$: The collection of all $\omega$-covers of a space $X$. An open cover $\mathcal U$ of $X$ is said to be $\omega$-cover if $X\notin\mathcal U$ and for each finite $F\subseteq X$ there exists a $U\in\mathcal U$ such that $F\subseteq U$.

1. A space $X$ is said to have the Menger property if for each sequence $(\mathcal U_n)$ of open covers of $X$ there exists a sequence $(\mathcal V_n)$ such that for each $n$ $\mathcal V_n$ is a finite subset of $\mathcal U_n$ and $\{\cup\mathcal V_n : n\in\mathbb N\}$ covers $X$.

2. A space $X$ is said to have the Scheepers property if for each sequence $(\mathcal U_n)$ of open covers of $X$ there exists a sequence $(\mathcal V_n)$ such that for each $n$ $\mathcal V_n$ is a finite subset of $\mathcal U_n$ and $\{\cup\mathcal V_n : n\in\mathbb N\}\in\Omega$ or $\cup\mathcal V_n=X$ for some $n$.

A space $X$ is said to be Menger (respectively, Scheepers) if $X$ has the Menger (respectively, Scheepers) property. It is well known that every Scheepers space is Menger but the coverse is not true (see here). We are looking for a ZFC example of a Menger space which is not Scheepers (here, it means without any additional set-theoretic assumption, is there a Menger space which is not Scheepers?).

Answer 1

It is consistent that every Menger space is Menger in all finite powers (at least for "nice" enough spaces), see the paper Products of Menger spaces in the Miller model (L. Zdomskyy, Advances in Mathematics 335 (2018), 170–179.)

Spaces with all finite powers Menger are $S_\text{fin}(\Omega,\Omega)$, and in particular Scheepers.

So, there is no ZFC example of a Menger space that is not Scheepers.