Every regular, scattered and Lindelöf space with countable pseudocharacter has countable cardinality.

Question

Let $(X, \tau)$ be a regular, scattered and Lindelöf space with countable pseudocharacter. Is it true that $X$ needs to be countable?

What I have tried (using the Theorem 2.2 from M. E. Gewand, “The Lindelöf degree of scattered spaces and their products,” Journal of the Australian Mathematical Society (Series A) 37 (1984), 98–105): Let $(X, \tau)$ be a regular scattered Lindelöf space with countable pseudocharacter.
(Before going on, $X_{\delta}$ denotes the topological space of $X$ with the topology generated by its $G_{\delta}$ sets under the topology $\tau$).

As a consequence of the Theorem 2.2 ($X$ is $T_3$, scattered and Lindelöf), $X_{\delta}$ is Lindelöf. Now, as each point of $X$ is a $G_{\delta}$ set, then $X_{\delta}$ is discrete and $\mathcal{A} = \{ \{x\} : x \in X \}$ is an open cover of $X_{\delta}$. Finally, as $X_{\delta}$ is Lindelöf, $X$ needs to be countable.

Answer 1

Here is a direct proof.

Lemma: Every Lindelöf locally countable space is countable.

This is easy to check (for example see Must scattered spaces with points $G_\delta$ be locally countable?).

Theorem: Every regular Lindelöf scattered space with each point a $G_\delta$-set is countable.

Proof: Suppose $X$ is regular Lindelöf scattered and each point is a $G_\delta$-set. Scattered spaces are $T_0$ (see here), and $T_0$ regular spaces are $T_3$. So $X$ is Hausdorff and $T_1$.

Let $B$ be the set of points $x\in X$ at which $X$ is locally countable (i.e., such that $x$ has a countable nbhd). The set $B$ is open in $X$. So its complement $A=X\setminus B$ is closed, hence Lindelöf. Suppose by contradiction that $A$ is not empty. By the scattered property, there is a point $x\in A$ that is isolated in $A$. So there is an open set $U\subseteq X$ with $U\cap A=\{x\}$.

Since the singleton $\{x\}$ is a $G_\delta$, we have $\{x\}=\bigcap_{n=1}^\infty G_n$ for some open sets $G_n\subseteq U$. By regularity of $X$ we can also assume $\overline G_1\subseteq U$.

Each of the $\overline G_1\setminus G_n$ is closed in $X$, hence Lindelöf. And it is also locally countable as a subset of the locally countable open set $U\setminus\{x\}\subseteq B$. Thus $\overline G_1\setminus G_n$ is countable by the Lemma. And since the set $G_1\setminus\{x\}$ is covered by these countably many countable sets, it is also countable. But then $G_1$ is a countable nbhd of $x$, which contradicts the fact that $x\in A$. Therefore $A=\emptyset$ and $X$ is locally countable. Finally, one more application of the Lemma shows that $X$ is countable.