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Take the free abelian group $G={\mathbb{Z}}^2$, whose rank of a group is two, and its finite-index subgroup $H=2\mathbb{Z}\times \mathbb{Z}$. The subgroup has index two and is again isomorphic to ${\mathbb{Z}}^2$, so its rank is two. The Nielsen–Schreier formula would instead give $1+2(2-1)=3$. Henceis a counterexample.

Let $r=d(G)$ be the rank of a group, let $S$ be a generating set of size $r$, and suppose $[G:H]=d$. Choose a Schreier transversal adapted to a spanning tree in the Schreier coset graph. By Schreier's lemma, $H$ is generated by the elementsThere are $dr$ candidates. The $d-1$ oriented edges in the spanning tree give trivial candidates, leaving at most $dr-(d-1)$. This proves the Schreier index-rank inequality