 Solution

On the unit three-ball $B^3$, the distance in the Poincare ball model is the infimum of the lengthsover piecewise smooth paths joining the two points. Equivalently the hyperbolic metric as a metric-space distance isIt has curvature-minus-one normalization; in particular $d(0,x)=2\mathrm{artanh}|x|$.

For a nonempty finite set $p_1,\dots ,p_m$, define $R(x)={\max }_{j}d(x,p_j)$. The triangle inequality gives $|R(x)-R(y)|\le d(x,y)$, so $R$ is continuous. Take $R_0=R(p_1)$. Any improving center satisfies $d(x,p_1)\le R(x)\le R_0$, and thus lies in the closed hyperbolic ball $K=\overline{B}(p_1,R_0)$.

This ball is compact: an isometry sending $p_1$ to the origin identifies it with the Euclidean closed ball of radius $\tanh (R_0/2)<1$, and the Euclidean and hyperbolic topologies agree inside $B^3$. The continuous function $R$ attains a minimum on $K$. Centers outside $K$ have $R>R_0$, so this is also the global minimum. Its associated closed ball proves a minimum enclosing ball in hyperbolic space exists. For the empty finite set, any center and radius zero suffice.