Let $S=conv(H \cup\{x\} )$ denote the convex hull of $H \cup\{x\}$ where $H \subset \Bbb{R}^n$ is a halfspace and $x\in\Bbb{R}^n, x\notin H$.
I need to prove that $S$ is not a polyhedron and my definition of a polyhedron is a finite intersection of halfspaces, ie $P$ is a halfspace if $P=\bigcap_{i=1}^{m}H_i$ for each $H_i$ a halfspace.
Is it sufficient to show that a single point is an infinite intersection of halfspaces or is there something else I am missing?
First, show that any element of $S$ can be written as a convex combination of $x$ with a single element of $H$.
Then, if you draw all rays from $x$ that intersect $H$, you will see that $S$ is basically the union of $\{x\}$ with an open half space whose boundary intersects $x$ and is parallel to that of $H$. Then $S$ is not closed (it contains points arbitrarily close to the boundary, but not the boundary), whereas polytopes must be closed by your definition.