I think there aren't any discrete paradoxical subsets in $\mathbb{R}^2$ (any isometry that mapped a discrete subset into itself would have to either be a glide-reflection, a translation or a rotation by $2\pi/n$, and hence the subgroup of the isometry group on $\mathbb{R}^2$ generated by isometries that map a discrete set into itself would have only translations and glide-reflections as elements of infinite order and thus no free subsemigroup of rank 2, hence the subgroup would be supramenable).

Say that a subset in a metric space is uniformly discrete if there is an $r>0$ such that every pair of points has distance greater than $r$.

Are there any uniformly discrete paradoxical subsets in $\mathbb{R}^3$?

By definition, given a group $G$ acting on a set $X$, a subset $E\subseteq X$ is $G$-paradoxical if $E\neq\emptyset$ and there are $n,m$ and pairwise disjoint subsets $U_1,\dots,U_{n+m}$ of $E$ such that $E=\bigcup_{i=1}^ng_iU_i=\bigcup_{i=n+1}^{n+m}g_iU_i$. That is, one can divide $E$ into pieces that can be reassembled into $E$ twice over. In the above question, $G$ is meant to be the isometry group of the Euclidean space.