= Solution
In the hyperbolic-ball model, the <limit set of a Kleinian group> is the boundary accumulation set of an orbit $Go$ of an interior point: $\Lambda(G)=\overline{Go}\cap\partial\mathbb H^3$. The choice of $o$ does not matter, since any two such orbits remain a fixed hyperbolic distance apart. It is closed and $G$-invariant. The complicated/non-elementary hypothesis excludes relatively compact groups and elementary groups with at most two limit points; in particular the limit set is nonempty. Having no finite boundary orbit alone would not exclude a compact rotation group, whose interior orbits are bounded.
If $\Lambda(G)$ were finite and nonempty, the orbit of any of its points would be a finite boundary orbit, contrary to the hypothesis. Thus it is infinite. In fact it cannot be countable. We give the essential minimality/perfectness argument. A sequence of <Möbius transformations> sending an interior point toward $\xi\in\Lambda$ has a subsequence converging on the boundary away from one exceptional point to the constant $\xi$. This follows by normalizing the representing $2\times2$ <matrices>: a diverging sequence has a rank-one projective limit, with one kernel line and one image line. For any boundary point $\eta$, the absence of a finite orbit allows choosing $h\in G$ with $h\eta$ different from that exceptional point; then $g_jh\eta\to\xi$. Hence the orbit of every $\eta\in\Lambda$ is dense in $\Lambda$.
If $\xi$ were an isolated limit point, density of every orbit would make every point belong to its orbit: each orbit must hit the open singleton $\{\xi\}$. Every point would then be isolated by equivariance, making the compact set $\Lambda$ discrete and therefore finite, a contradiction. Thus $\Lambda$ is a nonempty perfect compact metric space. Repeatedly choosing two disjoint smaller neighbourhoods meeting this perfect set, with diameters tending to zero, associates distinct nested limits to all infinite binary sequences. \b[The limit set is therefore uncountable.]
<Jorgensen inequality> states that for a non-elementary discrete group generated by lifts $A,B\in\operatorname{SL}(2,\mathbb C)$,
$$
|\operatorname{tr}^2 A-4|+|\operatorname{tr}(ABA^{-1}B^{-1})-2|\ge1.
$$
For a concrete failure of discreteness, take $A=\operatorname{diag}(e^{s/2},e^{-s/2})$ and $B=\bigl(\begin{smallmatrix}1&1\\-1&0\end{smallmatrix}\bigr)$ with $s>0$ small. The <fixed points> of $A$ and of $BAB^{-1}$ collectively include $0,\infty,-1$, so the group is non-elementary. Direct multiplication gives both summands equal to $4\sinh^2(s/2)$. Their sum is less than one when $s$ is sufficiently small, proving \b[this two-generator group is not discrete].
For a connected-limit-set example, the <modular group> generated by $z\mapsto z+1$ and $z\mapsto-1/z$ is discrete because its <matrices> have integer entries. It preserves the upper half-plane; its limit set is the extended real line, a circle. Indeed the orbit of the parabolic <fixed point> infinity consists of the rational points and infinity, dense in that circle. For a disconnected example, choose four pairwise disjoint round disks and two loxodromic maps pairing the exterior of each negative disk with the interior of its positive disk. The <Schottky group> generated by these maps is free and discrete by the disjoint-domain ping-pong construction. Successive reduced words give nested disjoint disks with shrinking diameters; their limiting points form a Cantor set. Thus its limit set is disconnected, unlike that of the <modular group>.
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