Working in ZFC, suppose we have the following sets. $C_0=\emptyset, C_{\alpha+1}$ is the set of all countable subsets of $C_\alpha$ and $C_\lambda= \cup_{\gamma<\lambda} C_\gamma$ for $\lambda$ a non-zero limit. Then I want to try to determine whether there is a $\beta$ with $C_\beta=C_{\beta+1}$.
I have had some ideas: inductively we can see that the rank of any set in $C_{\alpha}$ is a countable ordinal (countable union of countable sets). So any element of a $C_\alpha$ is in $V_{\omega_1}$ of the Von Neumann hierarchy. This set has some cardinality, $m$ say. Then by Hartogs' we have an ordinal that does not inject into $V_{\omega_1}$. Then using this, we see that for some $\gamma$ the $C_\delta$ with $\delta<\gamma$ are not distinct. Now it's not clear to me if we've reached a dead end in the proof. Because unless we can make some claim like that $C_\alpha\subset C_{\alpha+1}$ for each $\alpha$ which I can't easily see if that's true then we aren't done. (Maybe it's not even true that there exists such an $\beta$?)
Lemma. $C_\alpha\subseteq C_{\alpha+1}$ for all $\alpha$.
Proof. Let $S=\{\alpha:C_\beta\subseteq C_\alpha\text{ for all }\beta\lt\alpha\}$. Plainly $\lambda\in S$ if $\lambda$ is a limit ordinal or $0$. Suppose $\alpha\in S$; I have to show that $\alpha+1\in S$, i.e., that $C_\alpha\subseteq C_{\alpha+1}$. Suppose $x\in C_\alpha$; I have to show that $x\in C_{\alpha+1}$. Consider the least $\beta$ such that $x\in C_\beta$; so $\beta\le\alpha$ and $\beta$ is a successor ordinal, say $\beta=\gamma+1$. Then $x\subseteq C_\gamma\subseteq C_\alpha$, so (being countable) $x\in C_{\alpha+1}$, Q.E.D.
Corollary. $C_{\omega_1}=C_{\omega_1+1}$.
P,S. More elegantly, prove by induction that $C_\alpha$ is a transitive set of countable sets, i.e., every element of $C_\alpha$ is a countable subset of $C_\alpha$. If $C_\alpha$ is a transitive set of countable sets, it clearly follows that $C_\alpha\subseteq C_{\alpha+1}$, and that $C_{\alpha+1}$ is also a transitive set of countable sets.