Every finitely generated free group is subgroup separable

By [F1], choose a finite-index subgroup $K\le F$ with a free-product decomposition $K=H\ast J$. If $g\mathrm{\notin }K$, set $L=K$. Then $L$ has finite index in $F$, contains $H$, and omits $g$.

If $g\in K$, the free-product decomposition gives a retraction $r:K\to H$ that fixes $H$ and kills $J$. Since $g\mathrm{\notin }H$, the element $d=gr(g{)}^{-1}$ is nonidentity. By [F2] there is a homomorphism $\pi :K\to Q$ to a finite group with $\pi (d)\mathrm{\ne }1$. The homomorphism $k\mapsto (\pi (k),\pi (r(k)))$ to $Q\times Q$ has finite image. Let $L$ be the preimage of its diagonal subgroup. Then $L$ has finite index in $K$, contains $H$ because $r(h)=h$, and omits $g$ because $\pi (g)\mathrm{\ne }\pi (r(g))$. Thus $L$ also has finite index in $F$.

In either case let $N$ be the core of $L$ in $F$, the intersection of its finitely many conjugates. Then $N$ is a finite-index normal subgroup of $F$ contained in $L$. Consequently $H\subseteq HN\subseteq L$ while $g\mathrm{\notin }L$, so $g\mathrm{\notin }HN$. This is exactly the separation required by A subgroup is separable when it is closed in the profinite topology, and a group is LERF when every finitely generated subgroup is separable. Since $H$ and $g$ were arbitrary, $F$ is subgroup separable.