= Solution
Write $N=\Gamma'$. As a subgroup of a discrete group, it is a <Kleinian group>, even if it is not finitely generated. Since it is infinite, part (b) gives a nonempty closed <Kleinian limit set> $\Lambda(N)$. For $\gamma\in\Gamma$, conjugation and base-point independence give
$$
\gamma\Lambda(N)=\Lambda(\gamma N\gamma^{-1}).
$$
One can see this directly by applying $\gamma$ to $No$: the image is the orbit of $\gamma o$ under the conjugate subgroup, and the choice of interior point does not affect its limit set. Normality makes the conjugate subgroup equal to $N$, so $\Lambda(N)$ is $\Gamma$-invariant. Applying part (c) yields $\Lambda(\Gamma)\subseteq\Lambda(N)$.
Conversely $No\subseteq\Gamma o$, so every boundary accumulation point of the smaller orbit is one of the larger orbit: $\Lambda(N)\subseteq\Lambda(\Gamma)$. Therefore the <limit set of an infinite normal Kleinian subgroup> satisfies
$$
\boxed{\Lambda(\Gamma')=\Lambda(\Gamma).}
$$
Infinitude supplies nonemptiness, normality supplies invariance, and non-elementarity supplies minimality. Properness of the subgroup is not needed, and no finite-generation assumption on it has been inserted.
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