In the article AN EXAMPLE INVOLVING BAIRE SPACES (https://www.ams.org/journals/proc/1975-048-01/S0002-9939-1975-0362249-1/S0002-9939-1975-0362249-1.pdf) of H. E. White Jr. it is shown that, assuming CH, there exists a Baire space $Y$ such that $Y\times Y$ is not Baire.

For the construction of this space, is used the density topology on $\mathbb{R}$ (for details of this topology you can see https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-62/issue-1/The-density-topology/pjm/1102867878.full).

Denote by $\mathcal{T}$ the density topology on $\mathbb{R}$ and by $\mathcal{E}$ the Euclidean topology on $\mathbb{R}$.

The construction of the space $Y$ begins with an enumeration $(F_{\alpha})_{\alpha<\omega_{1}}$ of all $\mathcal{E}$-Borel sets of measure zero (Lebesgue measure on $\mathbb{R}$). Then, by transfinite recursion on $\omega_{1}$ is construct a sequence $(Y_{\alpha})_{\alpha<\omega_{1}}$ of countable rational vector subspaces of $\mathbb{R}$ and finally our space is $Y=\bigcup_{\alpha<\omega_{1}}Y_{\alpha}$.

My question is the following :

In the article it is mentioned that $Y$ is not extremally disconnected, does anyone have any idea how to prove that fact?

Remember that a topological space $X$ is extremally disconnected if the closure of every open subset of $X$ is open.

Thanks a lot.