There are two common proofs of "if $G$ is free abelian then any subgroup $H$ is also free abelian" when $G$ has infinite rank, one using Zorn's lemma as in Lang's Algebra, one using a well ordering of the basis (in a neat way!) as in Kaplansky's Set Theory and Metric Spaces. I am wondering if the following "brute force attempt" can be made to work.
First assume $G$ is of finite rank $n$. Fix a basis of $G$, and express every element of $H$ as a column vector. Arrange them into an "$n$ by $H$" matrix, denoted $W=(w_{ij})$. Consider the ideal generated by the entries in the first row. It has to be of form $(a)$ where $a=k_1w_{1j_1}+\cdots k_sw_{1j_s}$, $k_i$ integers and $w_{1j_i}$ some of the entries in the first row (it could be the zero ideal). Now on the left of $W$ create an empty $n\times n$ matrix $W'$, and write the vector $w=k_1w_{j_1}+\cdots+k_sw_{j_s}$ in the first column, where $w_{j_i}=(w_{1j_i},\dots,w_{nj_i})^T$. By subtracting suitable multiples of $w$ from (the infinitely many) columns of $W$, we can make all entries in the first row of $W$ zero. Repeat this $n$ times, $W$ will become a zero matrix, and the nonzero columns in $W'$ (they increase in the row index of the first nonzero entry) would be a basis for $H$.
I guess this is essentially the "column operation" part of Smith normal form algorithm. Now what to do if $G$ is of infinite rank? Given a well-ordered basis $E$ for $G$, we create an $E$ by $H$ matrix and try to repeat the above procedure transfinitely; at the $\omega$-th step I want to take the "limit" of the matrix $W$; it's not clear that this is well defined, or even if it is, whether some column can now have infinitely many nonzero entries, so they don't represent elements in $H$ any more.