Context: this is intended to justify the second paragraph of Ethan Bolker's answer to my previous question concerning convex polyhedrons.
In $\Bbb R^n$ (assming $\color{red}{n\ge 3}$), there are two $(n - 1)$-simplices $F,G$ neither of which is "coplanar" with $0$, i.e. $0$ is in the affine hull of neither. Also, we assume $F,G$ aren't coplanar themselves. Finally, there is a plane $\pi$ (or $(n-1)$-dim affine subspace) that separates $F,G$ from the $0$, i.e. $\pi$ cuts the whole space into two open regions one of which contains $F,G$ and the other contains $0$.
Show that, there exists an open set $U\in\Bbb R^n$, such that for each $n\in U, n\ne 0$, the plane $\{x\in\Bbb R^n\mid n^Tx=0\}$ intersects with both $F$ and $G$.
In 3D version: $F, G$ are any two triangles neither of which is coplanar with $0$. So it boils down to justifying an open set of normals whose corresponding subspaces intersect both $F,G$.
My idea: pick $f\in\text{relint} F$ and $g_{1,\cdots, n-2}\in\text{relint} G$ (in which relint means relative interior, i.e., points that belong to the simplex but aren't on its "boundary") in a way that $0, f,g_{1,\cdots,n-2}$ can indeed determine a $(n-1)$-dim subspace, whose normal vector we pick to be $n$. Then prove that for a sufficently small $\epsilon>0$, we will have $\{(n+v)^Tx=0\}$ still intersects $F,G$ for all $v\in B_{\epsilon}(0)$. But I don't know how to get my hands on it. Any help? Thanks in advance.
