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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, ${\dim }_{H}C\ge 1$, contradicting ${\dim }_{H}F<1$. Thus 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 $C=AB{A}^{-1}{B}^{-1}$. Conjugating the subgroup generated by $B,C$ by $M(z)=iz/(2-z)$ gives matricesThe product of the second matrix with the first is $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. 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 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 arxiv.org/pdf/math/0701572 .

A Tangent-disk Schottky group uses side-pairing disks with disjoint interiors but permits boundaries to meet. This convention must be distinguished from the strictly separated, purely loxodromic Schottky group convention. For example, the maps $A(z)=z+2$ and $B(z)=z/(iz+1)$ pair the half-plane disks $\Re z<-1,\Re z>1$ and the disks $|z-i|<1,|z+i|<1$. The ping-pong lemma gives a free discrete group. With $C=AB{A}^{-1}{B}^{-1}$, the subgroup generated by $B,C$ is conjugate by $M(z)=iz/(2-z)$ to the principal level-two modular group, generated by $\left(\begin{array}{cc}1& 0\\ 2& 1\end{array}\right)$ and $\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. Thus its Kleinian limit set contains the circle $|z-1|=1$ and its translate $|z-3|=1$. Its Hausdorff dimension is at least one and it cannot be homeomorphic to a single circle.