Is a finitely generated residually free group "almost LERF"? 
Let $G$ be a finitely generated residually free group.

(i.e. for each $1 \neq g \in G$ there exists a homomorphism $\tau \colon G \to F$ such that $F$ is a free group, and $\tau(g) \neq 1$.)

Let $H \lneq G$ be a proper finitely generated subgroup. Must there be
  a finite index proper subgroup $U$ of $G$ containing $H$?

If $G$ must be LERF then the answer is positive. Henry Wilton, in a beautiful work, has proved that limit groups (fully residually free groups) are LERF.
 A: (converted from the comments) No, $F_2\times F_2$ is a counterexample, where $F_2$ is free on 2 generators.
Recall that a group is LPF if the profinite closure of every f.g. subgroup of infinite index has infinite index. This fails if there is a profinitely dense f.g. subgroup.
[Remark: your property appeared in the 3-manifold literature as "$G$ has the engulfing property". It is an elementary remark (see page 10-11 here, where Property LPF is introduced) that a group has Property LPF iff each of its finite index subgroups has the engulfing property.] 
Now a way to get profinitely dense f.g. subgroups in $F_2\times F_2$ is as follows. Consider an aperiodic (= with no nontrivial finite quotient) infinite finitely presented group $P$ on 2 generators (or with $n$ generators, but then work with $F_n\times F_n$), and fix an epimorphism $f:F_2\to P$. Then the fibre product $$F_2\times_P F_2=\{(x,y)\in F_2\times F_2:f(x)=f(y)\}$$ is profinitely dense in $F_2\times F_2$ (because $P$ is aperiodic) and finitely presented (because $P$ is finitely presented) and has infinite index (because $P$ is infinite).
