A free group of rank at least two has subgroups of every finite rank

Choose a free basis $X$ of $F$ with at least two elements, and fix distinct $x,y\in X$. Let $L=⟨x,y⟩$. For any group $G$ and any function $u:\{x,y\}\to G$, extend $u$ to a function on $X$ by sending every basis element outside $\{x,y\}$ to $e_G$. The universal property in [L1] then gives a unique homomorphism $F\to G$, whose restriction to $L$ extends $u$. Hence $\{x,y\}$ is a free basis of $L$, so [L2] gives $\mathrm{rank}(L)=2$.

For $m=1$, the cyclic subgroup $⟨x⟩$ has free basis $\{x\}$. Now assume $m\ge 2$. Because $L$ is free on $\{x,y\}$, there is a surjective homomorphism $\pi :L\to (\mathbb{Z}\mathrm{/}(m-1),+)$ with $\pi (x)=1$ and $\pi (y)=0$. Let $H_m=\ker \pi$. Then $[L:H_m]=m-1$, so [L3] gives $\mathrm{rank}(H_m)=1+(m-1)=m$.

The subgroup $⟨x⟩$ handles $m=1$, and the subgroups $H_m\le L\le F$ handle every $m\ge 2$. Therefore $F$ has subgroups of every finite rank.