Every hereditarily Lindelöf scattered space is countable

Question

Fact: Every hereditarily Lindelöf scattered space is countable.

Here, a space is Lindelöf if every open cover of the space has a countable subcover, and a hereditarily Lindelöf space is one for which each of its subspaces is Lindelöf. A scattered space is one in which every nonempty subset $A$ has an isolated point in $A$.

The very interesting paper [1] (available here) studies some properties of Lindelöf scattered spaces. It mentions in its introduction the Fact above as a basic result in this area, apparently well-known. Can anyone provide a proof?

[1] Banakh, Taras; Brian, Will; Ríos Herrejón, Alejandro, First-countable Lindelöf scattered spaces, Topology Appl. 322, Article ID 108318, 29 p. (2022). ZBL1514.54003.

Answer 1

Yes, this is a well-known result. I guess it is contained in one of the books of Juhasz about cardinal functions. In fact, it holds for arbitrary cardinals, namely:

Let $X$ be an infinite, scattered topological space. Then $hL(X) = |X|$, where $hL(X) = \text{sup}\{L(A): A \subset X\}$ is the hereditarily Lindelöf degree of $X$.

The proof uses a well-known characterization of scattered spaces:

Definition
Let $\le$ be a well-ordering of the topological space $X$.
$\le$ is called right-separating, if for each $x \in X$, $X^{\le x} = \{y\in X: y\le x\}$ is open.
Then also $\{y\in X: y < x\} = \bigcup_{y < x} X^{\le y}$ is open.

Theorem
The topological space $X$ is scattered, if and only if there exists a right-separating well-ordering on $X$.

For the time being, I would prefer to omitt the proof, although I haven't found a (online) reference yet, only several places, where this result is mentioned. Please let me know, if further elaboration is needed.

Using the theorem, we can now prove the above statement:

"$\le$" is obvious.
"$\ge$": W.l.o.g. we may assume that $X = \lambda \in $ On and each $\alpha$, $\alpha < \lambda$, is open.
Assume that $hL(X) < |X| = |\lambda|$. Then $\kappa := hL(X)^+ \le |\lambda|$, and $\kappa$ is a regular cardinal. $\{\alpha: \alpha < \kappa\}$ is an open cover of $\kappa$. Hence there exists $T \subset \kappa$ with $\bigcup_{\alpha \in T} \alpha = \kappa$ and $|T| \le L(\kappa) \le hL(X)$. Hence $T$ is cofinal in $\kappa$, which implies $\kappa = cf(\kappa) \le |T| \le hL(X) < \kappa$.
Contradiction! Hence $|X| \le hL(X)$.

Answer 2

Here is my attempt after reasoning similarly to this answer: https://math.stackexchange.com/q/4954574, and making use of its Lemma.

Suppose $X$ is hereditarily Lindelöf and scattered.

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 $X$. Now, as $X$ is hereditarily Lindelöf, $B$ is Lindelöf. So, as $B$ is Lindelöf and locally countable, $B$ is countable.

Let $A=X\setminus B$. Suppose by contradiction that $A\neq \emptyset$. By the scattered property, $\exists x \in A$ such that there is an open set $U \subseteq X$ such that $U\cap A=\{x\}$. Since $x$ doesn't have any countable nbhds (it isn't an element of $B$), $U$ is necessarily uncountable. But $U \setminus \{x\} \subseteq B$ and $U \setminus \{x\}$ is uncountable, so this is a contradiction, as $B$ is countable. Therefore, $A = \emptyset$, and $X = B$, so $X$ is countable.