In this question we are only interested in convex polyhedra in the Euclidean space $\mathbb R^3$.
Polyhedra $P$ and $P'$ are said to be combinatorially equivalent iff there is a bijection between them (denoted here as $X\mapsto X'$) preserving the number of vertices, edges and faces and their relations (i.e. edge $E$ connects vertices $A$ and $B$ and separates faces $G$ and $H$ in $P$ iff edge $E'$ connects vertices $A'$ and $B'$ and separates faces $G'$ and $H'$ in $P'$). Note that we ignore possible chirality, and thus every polyhedron is combinatorially equivalent to its mirror image.
Recall that a subset $S$ of $\mathbb R^3$ is dense iff there is a point from $S$ in every neighborhood of every point of $\mathbb R^3$. For example, the set of all points with rational coordinates is dense.
Question: Is it true that for every dense subset $S$ and every polyhedron $P$ there is a combinatorially equivalent polyhedron whose all vertices belong to $S$?