Two finite linear subdivisions have a common simplicial refinement

Form the finite convex cell complex of all intersections $\sigma \cap \tau$, $\sigma \in K_1$, $\tau \in K_2$, and their faces. It covers the common polyhedron and every cell is contained in a simplex of each triangulation.

Choose a compatible simplicial triangulation $T$ of that finite cell complex. Each simplex of $T$ lies in an intersection cell, hence in one simplex of $K_1$ and one of $K_2$, and $\mathrm{\mid }T\mathrm{\mid }$ is their common underlying set. These are exactly the conditions for a common linear refinement. If the set is empty take the empty complex; for points and lower-dimensional intersections the same cell triangulation applies. Only finitely many interior-point choices are required.