Is there any example of a Lindelöf space that has no Menger dense subspaces? A space $X$ is said to be Menger if for each sequence $(\mathcal{U}_n)$ of open covers of $X$, there is a sequence $(\mathcal{V}_n)$ such that $\mathcal{V}_n$ is a finite subcollection of $\mathcal{U}_n$, $n\in\omega$, and $\{\bigcup\mathcal{V}_n:n\in\omega\}$ is an open cover of $X$. A space $X$ is Lindelöf if each open cover has a countable subcover.
My question was motivated from the fact that classical examples of Lindelöf spaces that are not Menger, have dense Menger subspaces.
 A: Every space $X$ with a dense Menger subspace must be weakly Menger, that is, for each sequence $\{\mathcal{U}_n: n<\omega \}$ of open covers there is a finite sub-collection $\mathcal{V}_n \subset \mathcal{U}_n$ such that $\bigcup \{\mathcal{V}_n: n <\omega \}$ is dense in $X$. Therefore it's enough to find a Lindelöf non-weakly Menger space.
Given a topological space $(Y, \tau)$ let us denote by $\mathcal{K}[Y]$ the space of all compact nowhere dense subsets of $Y$ endowed with the Pixley–Roy topology, that is, the topology generated by the family $\{[F,U]: F \in \mathcal{K}[Y], U \in \tau, F \subset U\}$ where $[F,U]=\{G \in \mathcal{K}[Y]: F \subset G \subset U\}$.
Let $\mathbb{P}$ be the space of irrationals and consider the space $Z=\mathcal{K}[\mathbb{P}]$. van Douwen, Tall and Weiss proved that $Z$ is a ccc non-separable first-countable zero-dimensional Baire space without isolated points (see Theorem 3 of van Douwen, Eric K.; Tall, Franklin D.; Weiss, William A. R., Non-metrizable hereditarily Lindelöf from CH, Proc. Am. Math. Soc. (to appear) ZBL0345.54016.) and therefore, by the Corollary from page 140 of the same paper, under CH, $Z$ contains a dense Luzin subspace $X$. Luzin means that every nowhere dense subset of $X$ is countable and it is easy to see that this feature, along with the ccc of $X$ implies that $X$ is hereditarily Lindelof. It remains to prove that $X$ is not weakly Menger, but since $X$ is dense in $Z$ it suffices to prove that $Z$ is not weakly Menger.
Indeed, since $\mathbb{P}$ is not Menger, there is a countable sequence $\{\mathcal{U}_n: n < \omega \}$ of open covers of $\mathbb{P}$ which witnesses that. For every $K \in Z$ let $\mathcal{U}^K_n$ be a finite subcollection of $\mathcal{U}_n$ which covers $K$. Then $\mathcal{O}_n=\{[K, \bigcup \mathcal{U}^K_n]: K \in Z \}$ is an open cover of $Z$ for every $n<\omega$. Let $\mathcal{G}_n$ be a finite subcollection of $\mathcal{O}_n$ and let $\mathcal{F}_n$ be the finite subset of $Z$ such that $K \in \mathcal{F}_n$ if and only if $[K, \bigcup \mathcal{U}^K_n] \in \mathcal{G}_n$. Then $\bigcup \{\mathcal{U}^K_n: K \in \mathcal{F}_n \}$ is a finite subcollection of $\mathcal{U}_n$, for every $n<\omega$, and therefore there is a point $y \in \mathbb{P} \setminus (\bigcup \{ \bigcup \{\mathcal{U}^K_n: K \in \mathcal{F}_n \}: n < \omega\})$. It follows that $[\{y\}, \mathbb{P}]$ is a non-empty open subset of $Z$ which is disjoint from $\bigcup \{\mathcal{G}_n : n < \omega \}$, thus showing that $Z$ is not weakly Menger.
