Is the Alexandroff double circle separable? Is the Alexandroff double circle separable (i.e. has a countable dense subset)?

The Alexandroff double circle is the space with underlying set $C = C_1 \cup C_2$, where $C_i$ is the circle of radius $i$ and centre $0$ in the complex plane. The basic open sets are:


$\{z\}$ for every $z \in \mathbb{C}$ with $|z| = 2$, and
$U \cup \{ 2z : z \in U\} \setminus F$, where $U$ is open in the normal topology of $C_1$ (i.e., a union of open arcs) and $F$ is a finite subset of $C_2$.


So far, I was able to show it is compact, Hausdorff, first countable but not second countable. Is there a way to connect this information to determining if it is separable or not?
No, it's not separable. The points of $C_2$ form an uncountably infinite discrete subspace. If there were some countable dense subset $S$, there would have to be an element of $S$ in each open set $\lbrace z \rbrace$ for all $z \in C_2$. Therefore, $C_2 \subseteq S$, which contradicts $S$ being countable.
No, it's not separable. The points of $C_2$ form an uncountably infinite discrete subspace. If there were some countable dense subset $S$, there would have to be an element of $S$ in each open set $\lbrace z \rbrace$ for all $z \in C_2$. Therefore, $C_2 \subseteq S$, which contradicts $S$ being countable.