Consider the action of $\mathbb{Z}^n$ on $\mathbb{R}^n$ by translation: $n.x= x+n$.
Is this action proper and discontinuous or no ?
We say that a group $G$ acts on a topological space $X$ properly and discontinuously if for every compact $K$ in $X$ and for every point $x \in K$, the set $G.x \cap K $ is finite.
To start, Let $K \subset \mathbb{R}^n$ be a compact set (it is a bounded and a closed set in $\mathbb{R}^n$ i.e $K$ is contained in some closed ball $B(C,R)$). Let $x=(x_1,...,x_n) \in K$ then $\mathbb{Z}^n.x = \lbrace (x_1+z_1,..., x_n+z_n), z_i \in \mathbb{Z}\rbrace $ and $\mathbb{Z}^n.x \cap K = \lbrace y \in K, y-x \in \mathbb{Z}^n \rbrace = (x+\mathbb{Z}^n) \cap K. $ It seems to me that it is a finite set but I don't know how to proceed to show this!
As you note, if $K\subset\Bbb{R}^n$ is a compact set then it is contained in some closed ball $B(C,R)$. Then $K-x$ is also compact, and it is also contained in some closed box $[-M,M]^n$. Then it follows that $$\big|(x+\Bbb{Z}^n)\cap K\big|=\big|\Bbb{Z}^n\cap(K-x)\big|\leq\big|\Bbb{Z}^n\cap[-M,M]^n\big|=(2M+1)^n.$$