= Solution
Suppose a <connected> subset $C\subset F$ contains distinct points $a,b$. The distance function $x\mapsto|x-a|$ is $1$-<Lipschitz>, and its image on $C$ is <connected> and contains $0$ and $|b-a|$. It therefore contains the whole interval between those values. A <Lipschitz map> cannot increase <Hausdorff dimension>: taking images of arbitrarily fine covers multiplies their diameter powers by at most a fixed constant. Since a nondegenerate interval has <Hausdorff dimension> one, $\dim_H C\ge1$, contradicting $\dim_HF<1$. Thus \b[every <connected> subset is a singleton], so $F$ is <totally disconnected>.
A <Möbius transformation> pairs two disks when it maps their boundary circles to one another and takes the exterior of the first disk onto the interior of the second. Two pairs of disks with disjoint interiors give the <ping-pong lemma> construction of a free two-generator group. Tangencies are allowed in the convention suggested by the question; this is often called a <Tangent-disk Schottky group>, rather than a strictly separated <Schottky group>.
Here is a concrete configuration that also makes the dimension argument precise. Take $A(z)=z+2$ and $B(z)=z/(iz+1)$. The first pairs the half-plane disks $\Re z<-1$ and $\Re z>1$, which are disks on the sphere tangent at infinity. The second pairs $|z-i|<1$ and $|z+i|<1$, tangent at zero; their interiors are also disjoint from the half-plane disks. These pairings give a free discrete <Kleinian group> by <ping-pong lemma>, with a nonempty open fundamental region.
Set $C=ABA^{-1}B^{-1}$. Conjugating the subgroup generated by $B,C$ by $M(z)=iz/(2-z)$ gives matrices
$$
MBM^{-1}=\begin{pmatrix}1&0\\2&1\end{pmatrix},
\qquad
MCM^{-1}=\begin{pmatrix}-3&2\\-2&1\end{pmatrix}.
$$
The product of the second <matrix> with the first is $\begin{pmatrix}1&2\\0&1\end{pmatrix}$. These are the standard parabolic generators of the principal level-two modular group, whose ideal-quadrilateral fundamental region tiles the upper half-plane and whose <Kleinian limit set> is the entire extended real line. Consequently the original group's <Kleinian limit set> contains the circle $|z-1|=1$, as well as its image under $A$, the distinct tangent circle $|z-3|=1$.
It follows that \b[the limit set has <Hausdorff dimension> at least one and is not <homeomorphic> to a circle]. Indeed a proper subset of a circle cannot itself be a circle: after deleting a missing point it lies in a line, which cannot contain an embedded circle. A space containing two distinct complete circles thus cannot be <homeomorphic> to a single circle. This explains why dimension at least one does not force circle topology. The tangent-circle terminology is consistent with the primary discussion of T-Schottky configurations in https://arxiv.org/pdf/math/0701572 .
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