According to 2.3 Theorem (p. 98) from Eklof and Mekler's Almost free modules, an abelian group $M$ is $\aleph_{1}$-free, that is, all its countable subgroups are abelian free, if and only if $M$ is torsion-free and every finite subset of $M$ is contained in a finitely-generated pure subgroup of $M$ (B is a pure subgroup of $M$ if $M/B$ is torsion-free). The left to right implication goes as follows.
$M$ is torsion-free since otherwise $M$ contains a finite torsion group, and thus cannot be $\aleph_{1}$-free. Let $S$ be a finite subset of $M$. If $\langle S\rangle_{\ast}$ is not finitely-generated, then there is a countably generated subgroup $N$ of $\langle S\rangle_{\ast}$ containing $S$ which is not finitely-generated. But then $N$ is not free, since it has finite rank but is not finitely-generated. This contradicts the assumption that $M$ is $\aleph_{1}$-free.
$\langle S\rangle$ denotes the group generated by $S$ and $\langle S\rangle_{\ast}$ denotes the pure closure of $\langle S\rangle$, that is, the smallest pure subgroup of $M$ containing $\langle S\rangle$, which has the form $\{x\in M:nx\in \langle S\rangle \text{ for some } n\in\mathbb{N}\}$.
What definition of rank are we using to conclude that $N$ has finite rank?
Best Answer
As suggested by OP, I'm promoting my comment to an answer.
The rank of a torsion-free abelian group 𝐴 is the dimension of the rational vector space $A\otimes_{\mathbb Z}\mathbb Q$. Equivalently, it is the maximum size of a linearly independent (over ℤ or over ℚ, they're equivalent) subset of 𝐴.