= Solution
A subgroup of the <Möbius group> $\operatorname{PSL}_2(\mathbb C)$ is a <discrete subgroup> when the identity is isolated in its induced topology, equivalently every group element is isolated. We prove the compact-set criterion for a <properly discontinuous group action>: for every compact $K\subset\mathbb H^3$, only finitely many $g\in G$ satisfy $gK\cap K\ne\varnothing$.
Fix an origin $o$ and take $K\subseteq\overline B(o,R)$. If $x\in K$ and $gx\in K$, the triangle inequality and invariance of hyperbolic distance give
$$
d(o,go)\leq d(o,gx)+d(gx,go)=d(o,gx)+d(x,o)\leq2R.
$$
The set of Möbius isometries with $d(o,go)\leq2R$ is compact. To see this explicitly, represent an <isometry> by a determinant-one complex matrix and use singular-value decomposition:
$$
g=k_1\begin{pmatrix}e^{s/2}&0\\0&e^{-s/2}\end{pmatrix}k_2,\qquad
k_1,k_2\in\operatorname{SU}(2),\quad s\geq0.
$$
The compact factors fix $o$, and the diagonal factor moves it hyperbolic distance $s$, as seen from the vertical dilation by $e^s$ in upper half-space. Thus the bounded-displacement set is the continuous image of $\operatorname{SU}(2)\times[0,2R]\times\operatorname{SU}(2)$, modulo central signs.
By the allowed closedness fact, its intersection with $G$ is a closed discrete subset of a compact space, hence finite. Otherwise an infinite sequence would accumulate in that intersection, contradicting isolation of its limit point. This proves \b[every discrete Möbius subgroup acts properly discontinuously on hyperbolic three-space]. Finite stabilizers are allowed; discreteness alone does not imply a free action.
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