convex polytope integer points is there a simple proof for the following lemma:
An unbounded convex polytope (defined by linear constraints) has either zero integer points or infinite many integer points.
 A: Edit: the following argument is true when the polyhedron $P$ is the intersection of finitely many linear constraints: $a_1 x_1 + \ldots + a_n x_n \leq b$.


*

*The coefficient of linear constraints must be rational. Otherwise, your lemma will not be true (e.g. consider $\sqrt{2}x_1-\sqrt{3}x_2=0$).

*Since your polyhedron $P \subset \mathbb{R}^n$ is unbounded and defined by rational linear constraints, it has a rational recession direction, i.e. 
there exists $ 0 \neq d \in \mathbb{Q}^n$ such that $x+\lambda d \in P $,
for all $x \in P$ and $ \lambda  \geq 0 $. 
Therefore, if $x_0\in P \cap \mathbb{Z}^n $ then $x_0 +\lambda d \in P \cap \mathbb{Z}^n $ for infinitely many values of $\lambda$.

A: Edit: Let me emphasize that I make the (obviously necessary) extra assumption that $P$ has non-empty interior. I am not assuming, however, that the bounding hyperplanes have rational normal directions.
This will (essentially) follow from this sublemma:
If $x$ is in the interior of $P$, then $P$ contains a $\delta$ neighborhood of a ray $x+te$, $t\ge 0$, for some direction $e\in S$.
To see this, consider $x+te$ for fixed $e$ and observe that if this intersects a bounding hyperplane $H$ for some $t_0>0$, then suitable points $x+t(e')e'$ will also be on $H$, with $t(e')\approx t_0$, for all $e'\approx e$ (that could only fail if $n_H\cdot e=0$ for the normal direction of $H$, but then we would have $x\in H$ already). So if every ray $x+te$, $e\in S$ intersected the boundary, then $P$ would be bounded after all by compactness of the unit sphere. Since there are only finitely many bounding hyperplanes, there is a fixed positive distance to all of them from my ray.
If we now have an $x\in \mathbb Z^n$ in the interior of $P$, then this tube $y+te$, $t\ge 0$, $|y-x|<\delta$ will contain infinitely many other integer points because $x+te$ at least gets arbitrarily close to $\mathbb Z^n$ (the details depend on the rational relations among the $e_j$).
If $x$ was on the boundary, a modified version of this still works.
