Let $G$ be an abelian group. 

The statement 

> If every subgroup of $G$ of finite rank is $\mathbf{Z}$-free, then $G$ is $\mathbf{Z}$-free.

is a theorem for $G$ countable, but false in general ($\mathbf{Z}^X$ for infinite $X$ is a counterexample).

[Recall that the rank, or $\mathbf{Q}$-rank of an abelian group $A$ is the maximal number of $\mathbf{Z}$-free elements, or equivalently the dimension over $\mathbf{Q}$ of $A\otimes_\mathbf{Z}\mathbf{Q}$. For instance $\mathbf{Q}$ and $\mathbf{Z}[1/n]$ have rank 1.]