I am reading a wonderful paper by Komjáth, "Three clouds may cover the plane," and am having difficulty proving that certain sets are countable.
Assume CH (the continuum hypothesis) holds. Let $a,b,c\in \mathbb{R}^2$ be three noncolinear points. Let $E$ be the set of all lines going through one of $a,b$ or $c$. Komjáth claims that, "by closure we can find the increasing, continuous, $\textbf{countable}$ decompositions,
$$E=\bigcup\{E_\alpha:\alpha<\omega_1\}$$ and $$\mathbb{R}^2=\bigcup\{X_\alpha:\alpha<\omega_1\}$$ such that
(1) $E_0=X_0=\emptyset$;
(2) $a,b,c\in X_1$;
(3) if $e_1,e_2\in E_\alpha$ then their intersection is in $X_\alpha$;
(4) if $p\in X_\alpha$, then the lines $pa,pb,pc$ are in $E_\alpha$."
Now, I am assuming that he implicitly assumes that we ordered $\mathbb{R}^2=\{r_\alpha:1<\alpha<\omega_1\}$ (by CH), and is using these two rules for the transfinite recursion (otherwise, I don't see how the recursion picks up): Let $\alpha>1$, then
(5) $r_\alpha\in X_\alpha$
(6) if $\alpha$ is a limit ordinal, then $X_\alpha = \bigcup_{\beta<\alpha}X_\beta$ and $E_\alpha =\bigcup_{\beta<\alpha}E_\beta$.
I want to know why the we the $X_\alpha$'s are countable. It is clear that, if $\alpha$ is a limit ordinal and $X_\beta$ is countable for $\beta<\alpha$ then $X_\alpha$ is countable (since $\alpha<\omega_1$), but I don't see how to prove that $X_\beta$'s are countable, for say finite $\beta<\omega$.
I considered maybe the downward Lowenheiem-Skolem theorem: Since $\mathbb{R}^2$ itself is such a model for $X_\beta$, and the rules for describing these sets is first order, then there must exist a countable model of $X_\beta$, so define $X_\beta$ as the smallest such model (intersect over them, the intersection must be nonempty). This seems promising to me, but I'm unsure whether my argument gives me models that are infact subsets of $\mathbb{R}^2$, or some strange set.
Am I missing some standard fact or construction I am unfamiliar with? Any help or advice would be appreciated.
(As a very minor point, I rather imagine that in fact he enumerated $\Bbb R=\{r_\alpha:\alpha<\omega_1\}$, so that there is an $r_0$.)
Suppose that $0<\eta<\omega_1$, and we’ve defined $E_\eta$ and $X_\eta$ in such a way that they are countable. Let $E_{\eta+1}(0)=E_\eta$ and $X_{\eta+1}(0)=X_\eta\cup\{r_\eta\}$. Given countable $E_{\eta+1}(n)$ and $X_{\eta+1}(n)$ for some $n\in\omega$, let $X_{\eta+1}(n+1)$ be $X_{\eta+1}(n)$ together with all points of intersection of lines in $E_{\eta+1}(n)$; $E_{\eta+1}(n)$ is countable, so there are only countably many points of intersection added to $X_{\eta+1}(n)$ to form $X_{\eta+1}(n+1)$, which is therefore still countable. Then let $E_{\eta+1}(n+1)$ be $E_{\eta+1}(n)$ together with all of the lines $pa,pb$, and $pc$ for $p\in X_{\eta+1}(n+1)$; $X_{\eta+1}(n+1)$ is countable, so that adds only countably many lines to $E_{\eta+1}(n)$, and $E_{\eta+1}(n+1)$ is still countable. Thus, we can continue the construction to $\omega$. Finally, let
$$X_{\eta+1}=\bigcup_{n\in\omega}X_{\eta+1}(n)$$
and
$$E_{\eta+1}=\bigcup_{n\in\omega}E_{\eta+1}(n)\;;$$
each of these is the union of countably many countable sets, so each is countable, and it’s clear that conditions (3) and (4) are satisfied at $\alpha=\eta+1$.