How do you prove that each face of a polytope is also a polytope? I think it may be through induction.
If you know that "polytopes" and "bounded polyhedra" are the same objects, then what you are asking is easy: It is straightforward that every face of a polyhedron is a polyhedron. Hence, every face of a bounded polyhedron is a bounded polyhedron.
Here by polytope I mean (as you) the convex hull of finitely many points. By a (convex) polyhedron I mean a subset of $\mathbb R^d$ defined by finitely many affine-linear inequalities. By a face of a polytope or polyhedron $P$ I mean the part of $P$ where some linear functional is maximized. (This is almost what you say in "the face, F, is the set x in K such that inner product $<x, u>$ is equal to an alpha", except one needs to assume that alpha is the maximum value taken by $<x, u>$ on $P$)