Problem) Show that every polytope is a bounded polyhedron.
Let $C=conv(V)$ be a polytope with a set $V$ that satisfies $|V|<\infty$. Let $C^* :=\{ y \mid <y,x>\leq 1 \text{ for all } x \in C \}$. Let $D_0(v)^-:=\{ x \mid <x,v>\leq1\}$ for $v \in V$. Let $y \in C^*$, then $<y,x>\leq 1 $ for all $x \in C$, it means that $<y,v>\leq 1 $ for all v $\in V \subset C$. Let $u\in \cap_{v\in V} D_0 (v)^-$, then $u\in D_0 (v)^-$ for all $v$, it means that $<u,v>\leq 1 $ for all $v \in V $, and $<u, \sum_{i\in I} t_i v_i> = \sum_{i\in I} t_i< u,v_i > \leq \sum_{i\in I} t_i \leq 1$ for any index set $I$ with $|I|<\infty$ where $v_i \in V$, $t_i\geq 0$, and $\sum_{i\in I} t_i=1$.
Then, $C^* = \cap_{v\in V} D_0(v)^-$.
It can be assumed that $0$ is in an interior ( if $0$ is not in an interior of $C$, then $C$ can be shifted so that $int(C)$ contains $0$.) Since $0\in int(C)\subset C$, there exists $\epsilon \in (0,1)$ such that $x$ with $\lVert x\rVert_2 \leq \epsilon $ satisfies that $x \in C$. Let $y \in C^* $, then $<y,x>\leq 1$ for all $x \in C$, $<y,x>\leq 1 $ for all $x$ with $\lVert x \rVert_2 \leq \epsilon$, $<y, (y/\lVert y\rVert_2) \epsilon > = \epsilon <1 \leq 1$, it means that $\lVert y \rVert_2 \leq 1/\epsilon$, so $C^*$ is bounded. Since $|V|$ is finite, $C^* =\cap_{v\in V} D_0(v)^-$ is a bounded polyhedron. I claim that $C^{**}=C$. Let $x \in C$ and $y\in C^*$, then $<y,x> \leq 1$, so $x \in C^{**}$. Let $x \notin C$.
By separating hyperplane theorem, there exist $a \in \mathbb{R}^d$ and $b\in \mathbb{R}$ such that $<a,x> > b > <a,z> $ for all $z \in C$. Due to the assumption $0\in int(C) \subset C$, $b>0$ and $<a/b,x>>1<a/b,z>$ for all $z \in C$.
Therefore $a/b \in C^*$, but $<a/b,x>>1$ so $x \notin C^{**}$. So $C=C^{**}$.
I claim that every bounded polyhedron is a polytope. I prove it by induction on dimension $d$.
It is trivial when $d=1$.
I suppose that the claim is true for $d\geq 2$.
Let $P=\cap_{\gamma \in \Gamma} \gamma $ be a intersection of finitely many closed half-spaces in $\mathbb{R}^d$ where $\gamma$ is a half space, and $|\Gamma|<\infty$.
Let $F_\gamma :=P \cap \partial \gamma $.
$F_\gamma $ is a bounded polyhedron of dimension at most $d-1$, $F_\gamma = conv(V_\gamma)$ with $V_\gamma \subset F_\gamma $ with $|V_\gamma|<\infty$ by induction hypothesis.
I claim that $P:=conv(V)$ where $V=\cup_{\gamma \in \Gamma} V_\gamma$.
$conv(V) \subset P$ is trivial by definition of $V$.
Let $x$ be a given and $l$ be a line s.t $x \in l$.
There exists $\alpha, \beta \in \Gamma $ s.t $y\in F_\alpha \cap l \cap p$ and $z \in F_\beta \cap l \cap P$, that is $x $ is in line segment between $y$ and $z$ so $y\in conv(V_\alpha)$ and $z \in conv(V_\beta)$. Therefore, $x\in conv(V_\alpha \cup V_\beta)\subset conv(V)$.
Since $C^*$ is a bounded polyhedron, it is a polytope.
Therefore, $C=C^{**}$ is a bounded polytope.