Let $\kappa$ be a regular uncountable cardinal. Let $A$ be a free group with basis $X$ and let $(C_\alpha)_{\alpha<\kappa}$ be an increasing sequence of subgroups of cardinality less than $\kappa$ such that if $\lambda<\kappa$ is a limit, then $C_\lambda=\bigcup_{\alpha<\lambda} C_\alpha$. Assume that $A=\bigcup_{\alpha<\kappa} C_\alpha$.
For any subset $S$ of $A$, let $<S>$ be the subgroup generated by $S$. Show that $\{\alpha<\kappa\mid C_\alpha=<C_\alpha\cap X>\}$ is unbounded in $\kappa$.
This is supposedly "easy to see," so Hodges does not give the proof as part of the proof of Lemma 1.3 of "In Singular Cardinality, Locally Free Algebras Are Free" [Algebra Universalis 12 (1981), page 208].
For each $\alpha<\kappa$, each $c\in C_\alpha$ is in the subgroup generated by a finite subset $X_c$ of $X$. The set $\bigcup_{c\in C_\alpha}X_c$ has cardinality less than $\kappa$, so by regularity there is some successor ordinal $C_\beta$ with $\alpha<\beta$ containing it.
We can repeat this to get a countable chain $C_\alpha\subset C_\beta\subset C_\gamma\subset\dots$. Since $\kappa$ is uncountable and regular, the supremum of the indices is a limit $\lambda$ less than $\kappa$. For every $c\in C_\lambda$, by continuity, there exists $\alpha<\lambda$ such that $c\in C_\alpha$, and hence there exists $\delta$ such that $\alpha<\delta<\lambda$ and $X_c\subseteq C_\delta\subseteq C_\lambda$.
Thus $C_\lambda\subseteq <C_\lambda\cap X>$ and the result follows.