Second countable vs. $G_\delta$-diagonal Here A Question on a second countable $T_2$ space, Paul asked if every second countable Hausdorff space has a $G_\delta$-diagonal. In the comments Brian M. Scott answered that, at the time (2015), the answer to that question might not be known. Interestingly enough, here Does second countable and functionally Hausdorff imply submetrizable?, Taras Banakh showed two years later that, as a consequence of a more general result, every second countable functionally Hausdorff space is submetrizable (in particular, every such space has a $G_\delta$-diagonal).
My question is whether the answer to Paul's original question is already known. I couldn't find anything definitive about it.
 A: There exists a counterexample to this question of Paul.
It suffices to find a second-countable Hausdorff space $X$ that has two properties:
(1) the space $X\times X$ is Baire;
(2) for any nonempty open sets $U,V\subseteq X$ we have $\overline U\cap\overline V\ne\emptyset$.
Such a space $X$ cannot have $G_\delta$-diagonal. Indeed, assuming that the diagonal $\Delta_X$ is of type $G_\delta$ in $X\times X$, we can write $(X\times X)\setminus \Delta_X$ as the union $\bigcup_{n\in\omega}F_n$ of closed subsets $F_n$ of $X\times X$. Property (2) implies that $X$ has no isolated points and hence the diagonal $\Delta_X$ is nowhere dense in $X\times X$. Since the space $X\times X$ is Baire, there exists $n\in\omega$ such that $F_n$ has non-empty interior in $X\times X$. Then there are nonempty open sets $U,V$ in $X$ such that $U\times V\subseteq F_n$ and hence $\overline U\times\overline V\subseteq F_n$. Then $(\overline U\times \overline V)\cap\Delta_X=\emptyset$ and hence $\overline U\cap\overline V=\emptyset$, which contradicts the property (2).
For the space $X$ one can take the projective space of the countable product of lines $\mathbb R^\omega$. So, $X$ is the quotient space of $\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$. Observe that the quotient map $q:\mathbb R^\omega\setminus\{0\}^\omega\to X$ is open. This fact can be used to show that $X$ is Hausdorff and its square $X\times X$ is Baire. Moreover, for any nonempty open set $U\subseteq X$ the closure $\overline{U}$ contains the image $q[(\{0\}^n\times \mathbb R^{\omega\setminus n})\setminus\{0\}^\omega]$ for some $n\in\omega$, which implies that the condition (2) is satisfied.
