Example of an uncountable scattered space with some properties This might be an easy question, maybe the example I'm looking for is common knowledge. As always, recall that a topological space $X$ is scattered if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. It is known that any first countable, $T_3$, Lindelöf and scattered space is countable. Is there an example of an uncountable, first countable, Hausdorff, Lindelöf and scattered space?
 A: The answer to the question is yes: $\mathsf{ZFC}$ alone proves there is an example of an uncountable, first countable, Hausdorff, Lindelöf, scattered space.
The example I will describe has scattered height $\omega$. By the second part of Taras' answer, having infinite scattered height is a necessary feature of the example, because we are not assuming $\mathfrak{b} = \aleph_1$.
Recall that a Bernstein set is a subset $X$ of $\mathbb R$ such that both $X$ and its complement have nonempty intersection with every uncountable closed subset of $\mathbb R$. Such subsets of $\mathbb R$ can be proved to exist using the Axiom of Choice. In fact, the standard proof of the existence of Bernstein sets can be modified slightly to get a partition of $\mathbb R$ into infinitely many Bernstein sets:
Lemma: There is a partition of $\mathbb R$ into a countably infinite number of sets, such that each of these sets has nonempty intersection with every uncountable closed subset of $\mathbb R$.
(Actually, it is possible to partition $\mathbb R$ into $\mathfrak{c}$ Bernstein sets. But to construct our example, we want a partition into just $\aleph_0$ Bernstein sets.)
Let $X_0, X_1, X_2, X_3, \dots$ be a countable collection of pairwise disjoint Bernstein sets, as described in the lemma.
The underlying set of our example is $\mathbb R$, endowed with the topology generated by the following basic open sets. Given $x \in \mathbb R$, there is a unique $n$ such that $x \in X_n$; we take the basic open neighborhoods of $x$ to be sets of the form $\{x\} \cup \left( U \cap \bigcup_{i < n}X_i \right)$, where $U$ is a rational open interval in $\mathbb R$ containing $x$ (i.e., $U$ is a basic open neighborhood of $x$ in the normal Euclidean topology on $\mathbb R$).
This space is clearly uncountable, because its underlying set is $\mathbb R$. It is first countable, because the neighborhood bases described above are all countable. This space is Hausdorff because rational Euclidean-open intervals are still open in this topology (i.e., it refines the usual topology on $\mathbb R$). It is scattered because for any $Y \subseteq \mathbb R$, there is some minimal $n$ such that $Y \cap X_n \neq \emptyset$, and any point of $Y \cap X_n$ is isolated in $Y$.
Finally, it remains to show this topology is Lindelöf. Roughly, the idea is that for each $n$, we can use countably many of the neighborhoods of the points on level $n$ of our space (i.e., points in $X_n$) to reach down and cover all but countably many of the points on lower levels.
Suppose $\mathcal U$ is an open cover for this space. By shrinking the elements of $\mathcal U$ if necessary, we may (and do) assume that each member of $\mathcal U$ is a basic open set as described above. For each $n$, let $\mathcal U_n$ denote the set of all those $U \in \mathcal U$ such that $|U \cap X_n| = 1$; and for each $U \in \mathcal U_n$, let $I_U$ denote the (unique) rational Euclidean-open interval such that $U \cap \bigcup_{i < n}X_i = I_U \cap \bigcup_{i < n}X_i$.
Fix $n \in \omega$ with $n > 0$. For each Euclidean-open interval $I \in \{ I_U :\, U \in \mathcal U_n \}$, choose some particular $U(I) \in \mathcal U_n$ such that $I_{U(I)} = I$.  The set of all rational Euclidean-open $I$'s is countable, so the set of all such $U(I)$'s is a countable subset of $\mathcal U_n$. Let us denote this countable subset of $\mathcal U_n$ by $\mathcal V_n$. Now observe that $W_n = \bigcup \{ I(U) :\, U \in \mathcal V_n \}$ is a Euclidean-open subset of $\mathbb R$ containing $X_n$. Because $X_n$ is a Bernstein set, this means $\mathbb R \setminus W_n$ is countable. In particular, this implies that $\bigcup \mathcal V_n$ contains all but countably many points of $\bigcup_{i < n}X_i$.
Now unfix $n$. By the previous paragraph, $\bigcup_{n \in \omega \setminus \{0\}} \mathcal V_n$ is a countable subset of $\mathcal U$ that covers all but countably many points of our space. Adding in one more open set for each of the points not covered already by these sets, we obtain a countable subcover of $\mathcal U$.
A: There exists an example for this question under Continuum Hypothesis, more precisely, under the assumption $\mathfrak b=\omega_1$. In this case by Theorem 10.2 in the van Douwen's survey paper ``The integers and Topology'', there exists an uncountable set $C\subseteq\mathbb R\setminus\mathbb Q$, which is concentrated at rationals in the sense that for every open set $U\subseteq\mathbb R$ with $\mathbb Q\subseteq U$ we the complement $C\setminus U$ is countable.
Consider the uncountable space $X=\mathbb Q\cup C$ endowed with the topology $\tau$ consisting of the sets $W\subseteq X$ such that for every $q\in W\cap\mathbb Q$ there exists $\varepsilon>0$ such that $\{x\in C:|x-q|<\varepsilon\}\subseteq W$.
It is easy to see that the space $(X,\tau)$ is (functionally) Hausdorff, first-countable, Lindelof and scattered. More precisely, the subspace $C=X\setminus\mathbb Q$ is open and discrete and $\mathbb Q$ is closed and discrete in $X$. So, $X$ is scattered of finite scattered height.
To see that $X$ is Lindelof, observe that for every open cover $\mathcal U$ there exists a countable subfamily $\mathcal V\subseteq\mathcal U$ such that $\mathbb Q\subseteq \bigcup\mathcal V$. Since $C$ is concentrated at $\mathbb Q$, the difference $C\setminus\bigcup\mathcal V$ is countable and hence we can choose a countable subfamily $\mathcal V'\subseteq \mathcal U$ such that $C\setminus \bigcup\mathcal V\subseteq\bigcup\mathcal V'$. Then $\mathcal V\cup\mathcal V'$ is a required countable subcover of $\mathcal U$, witnessing that $X$ is Lindelof.

The above example cannot be constructed in ZFC because of the following
Theorem. Under $\mathfrak b>\omega_1$, every scattered Lindelof $T_1$-space of countable pseudocharacter and finite scattered height is countable.
Prior writing down the proof of this theorem, let us recall the definition of the scattered height.
Given a subset $A$ of a topological space $X$, let $A^{(1)}$ be the set of nonisolated points of $A$. Observe that for a closed subset $A$ of $X$, the set $A^{(1)}$ is closed in $A$. Let $X^{(0)}=X$ and for every nonzero ordinal $\alpha$ let
$$X^{(\alpha)}=\bigcap_{\beta<\alpha}\big(X^{(\beta)}\big)^{(1)}.$$
By transfinite induction it can be shown that $(X^{(\alpha)})_\alpha$ is a decreasing sequence of closed subsets of $X$, so it should stabilize at some ordinal $\alpha$. The smallest ordinal $\alpha$ such that $X^{(\alpha+1)}=X^{(\alpha)}$ is called the scattered height of $X$ and is denoted by $\hbar(X)$. A topological space $X$ is scattered if and only if $X^{(\hbar(X))}=\emptyset$.
Proof of Theorem. Assume that $\mathfrak b>\omega_1$. By Mathematical Induction we shall prove that for every natural number $n$, every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)\le n$ is countable.
For $n=0$, each scattered space $X$ with $\hbar(X)\le 0$ is empty and hence countable. Assume that for some positive integer number $n$,  every Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)<n$ is countable. Take a Lindelof scattered space $X$ of countable pseudocharacter and scattered height $\hbar(X)=n$. Then $X^{(n)}=\emptyset$ and $X^{(n-1)}$ is a nonempty discrete subspace of $X$. Being a closed  subspace of the Lindelof space  $X$, the discrete space $X^{(n-1)}$ is Lindelof and countable. Write $X^{(n-1)}$ as $\{x_k\}_{k\in\omega}$. Since $X$ has countable pseudocharacter, for every $k\in\omega$ there exists a decreasing sequence $(U_{k,m})_{m\in\omega}$ of open sets in $X$ such that $\bigcup_{m\in\omega}U_{k,m}=\{x_k\}$. Assuming that the space $X$ is uncountable, choose a transfinite sequence $\{y_\alpha\}_{\alpha\in\omega_1}$ consisting of pairwise disjoint elements of $X\setminus X^{(n-1)}$. For every $\alpha\in\omega_1$, let $f_\alpha:\omega\to\omega$ be the function assigning to each $k\in\omega$ the smallest number $f_\alpha(k)$ such that $y_\alpha\notin U_{k,f_\alpha(k)}$.  The number $f_\alpha(k)$ exists since $y_\alpha\notin\{x_k\}=\bigcap_{m\in\omega}U_{k,m}$. Since $\mathfrak b>\omega_1$, there exists a countable subfamily $\mathcal F\subseteq\omega^\omega$ such that for every $\alpha\in\omega_1$ there exists a function $f\in\mathcal F$ such that $f_\alpha\le f$. For every $f\in\mathcal F$, consider the open neighborhood $U_f=\bigcup_{k\in\omega}U_{k,f(k)}$ of the set $\{x_k\}_{k\in\omega}=X^{(n-1)}$. The choice of $\mathcal F$ guarantees that $\{y_\alpha\}_{\alpha\in\omega_1}\cap\bigcap_{f\in\mathcal F}U_f=\emptyset$. By the Pigeonhole Principle, for some $f\in\mathcal F$, the set $\Omega=\{\alpha\in\omega_1:y_\alpha\notin U_f\}$ is uncountable. On the other hand, $Y=X\setminus U_f$ is a closed (and hence Lindelof) subspace of $X$.  It follows from $Y\cap X^{(n-1)}=\emptyset$ that $\hbar(Y)<n$ and hence $Y$ is countable, by the inductive hypothesis. But this contradicts the uncountability of the set $\{y_\alpha\}_{\alpha\in\Omega}\subseteq Y$.

Can the theorem be generalized to scattered spaces of arbitrary scattered height?
Problem. Assume that $\mathfrak b>\omega_1$. Is every scattered Lindelof $T_1$-space of countable pseudocharacter countable?

Remark. Thanks to the intervention of Will Brian, I realized that the  equality $\mathfrak d=\omega_1$ in the initial redaction of my answer can be replaced by the weaker equality $\mathfrak b=\omega_1$. So, now we have the following characterization.
Theorem. The following conditions are equivalent:

*

*$\mathfrak b>\omega_1$.


*Every Lindelof scattered $T_1$-space of countable pseudocharacter finite scattered height is countable.


*Every first-countable Lindelof scattered Hausdorff space of scattered height 2 is countable.
The answer of Will Brian shows that this theorem does not extend to scattered Lindelof spaces of infinite scattered height.
