Could someone explain the underlined part from Introduction to Stochastic Programming book (2nd Edt.) of JR Birge -F Louveaux? 
Questions:
Q1- Why only finitely many different such hyperplanes?
Q2- How does this complete the proof?
 A: It seems that the proof tries to establish that $K_2(\xi)$ has a finite number of defining linear inequalities, therefore, is a polyhedron. However, it's not clear to me that the argument is entirely correct.
First, it may not be possible to find an $x$ and $\xi$ s.t. $x$ is not in $K_2(\xi)$. For example, if $W(\omega)=[-1, 1]$ for all $\omega$ and $y\in\mathbb{R}^2$, then $\text{pos}W(\omega)=\{t\,|\,W(\omega)y=t, y=[y_1,y_2]^T\ge 0\}=\{t\, |\, t = y_1-y_2, \, y_1\ge 0, y_2\ge 0\}=\mathbb{R}$. So no such $x$ and $\xi$ exists in the assumption in the beginning of the proof. Of course, if no such $x$ exists, then $K_2(\xi)$ is the entire space, so it's a polyhedron. However, the proof does not seem to consider this possibility.
Second, there could be infinitely many separating hyperplanes that pass through the origin and separate a point from a convex cone. It doesn't seem to make sense to claim there are only finitely many such hyperplanes, if only separation argument is invoked.
Claim a) directly follows from the fact: ''the projection of a polyhedron is a polyhedron.'' Because $K_2(\xi)$ is the projection of the following polyhedron $\{(x,y) \,|\, y\ge 0, W(\omega)y=h(\omega)-T(\omega)x\}$ to the space of $x$. This fact can be proved by the Fourier-Motzkin elimination (see e.g. Projection of a polyhedron is a polyhedron).
