TITLE: Example of an uncountable scattered space with some properties
QUESTION [9 upvotes]: 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?

REPLY [11 votes]: 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 $\mathcal U$.