Every finitely generated subgroup of a finite-rank free group is a free factor of a finite-index subgroup

Choose a finite free basis $X$ of $F$ and generators $h_1,\dots ,h_m$ of $H$. At a base vertex $v$, attach one reduced loop labeled by each $h_i$. Repeatedly fold pairs of equally labeled edges with the same initial vertex. Each fold preserves the set of labels of closed based paths and strictly decreases the number of edges, so after finitely many folds we obtain a finite connected folded pointed $X$-labeled graph $\mathrm{\Gamma }$ whose closed based labels are exactly the subgroup $H$.

For a fixed $x\in X$, each existing $x$-edge contributes one outgoing $x$-incidence and one incoming $x$-incidence. Therefore the number of vertices missing an outgoing $x$-edge equals the number missing an incoming $x$-edge. Pair those deficits and add finitely many new $x$-edges; doing this for every $x\in X$ produces a finite connected folded $X$-regular graph $\widehat{\mathrm{\Gamma }}$ containing $\mathrm{\Gamma }$.

Choose a spanning tree $T_{\mathrm{\Gamma }}$ of $\mathrm{\Gamma }$ and extend it to a spanning tree $T$ of $\widehat{\mathrm{\Gamma }}$. For each oriented edge $e$ outside $T_{\mathrm{\Gamma }}$, the tree path from the basepoint to the initial vertex of $e$, followed by $e$ and the reverse tree path from its terminal vertex, is a based loop. The loops obtained from one orientation of every non-tree edge freely generate the based loops of $\mathrm{\Gamma }$: deleting tree backtracking rewrites every closed path in them, while a nonempty reduced word in these loop generators leaves a non-tree edge after cancellation and hence is a nontrivial reduced path. Since $\mathrm{\Gamma }$ is folded and its closed labels are exactly $H$, their labels form a free basis $B_H$ of $H$.

Let $K$ be the set of labels of closed based paths in $\widehat{\mathrm{\Gamma }}$. The graph is folded and $X$-regular, so every reduced word on $X\bigsqcup X^{-1}$ is read from the basepoint along a unique path. Two words end at the same vertex exactly when their quotient labels a closed based path, that is, exactly when they lie in the same right coset of $K$. Hence the vertices of $\widehat{\mathrm{\Gamma }}$ are the right cosets of $K$, so $[F:K]=\mathrm{\mid }V(\widehat{\mathrm{\Gamma }})\mathrm{\mid }$ is finite. Because every closed based path of $\mathrm{\Gamma }$ is still closed in $\widehat{\mathrm{\Gamma }}$, one has $H\le K$.

Apply the same tree-loop argument to the finite $X$-regular graph $\widehat{\mathrm{\Gamma }}$. Its non-tree edges consist of those of $\mathrm{\Gamma }$ together with a disjoint set of added edges, so their loop labels form a free basis $B_H\bigsqcup C$ of $K$. By [L3], the subgroup generated by $B_H$ is a free factor of the subgroup generated by $B_H\bigsqcup C$, namely of $K$. Thus $H$ is a free factor of the finite-index subgroup $K$.