Everything is in $\mathbb R^n$.
Consider two sets $A, B$ such that they are disjoint, convex, positive cones. That is, $\forall k > 0, kA = A, kB = B$.
Also assume that $B$ is closed and polygonal, and $A$ is relatively open.
I think it is true that there exists some hyperplane that separates them, that is strict on at least one side. That is, there exists some $p\neq 0$ such that $p \cdot A > 0, p \cdot B \leq 0$.
How to prove it?
Consider a minimal counterexample in $\mathbb R^3$. $B$ is the closed cone $\{(x,y, z): x^2 + y^2 \leq z^2, z \geq 0\}$, and $A$ is the positive cone of $\{(x,y,z):x = z, y\in (0, 1), z > 0\}$.
The obvious fix is to make $A$ open, but this is just a general result: if $A, B$ are disjoint and convex, $A$ open and $B$ closed, then there exists such a separation. The "conic" property is entirely unused.
If we also assume $B$ is polygonal, then perhaps something like "enumerate all facets" may work.