Let $(G_\alpha)$ and $(K_\alpha)$ $(\alpha<\aleph_1)$ be strictly increasing chains of countable sets such that if $\alpha$ is a limit, then $G_\alpha=\bigcup_{\beta<\alpha}G_\beta$ and $K_\alpha=\bigcup_{\beta<\alpha}K_\beta$. Assume $\bigcup_{\alpha<\aleph_1}G_\alpha=\bigcup_{\alpha<\aleph_1}K_\alpha$. Does there exist a club $C$ such that $K_\gamma=G_\gamma$ for all $\gamma\in C$?

In the paper ``The Abelianization of Almost Free Groups" (the end of the proof of Lemma 2.5 on page 1801), this assertion is made where the $G_\alpha$ and $K_\alpha$ are subgroups of a free group with some additional properties, but the author makes the assertion I want without any explanation, so I assume the reason must be simple and may not depend on the group theory.

club setis a subset of a limit ordinal which is closed under the order topology, and is unbounded (see below) relative to the limit ordinal. The nameclubis a contraction of "closed and unbounded". $\endgroup$ – YCor Jul 6 '18 at 7:59