Proposition: Suppose a group $\Gamma$ acts properly by isometries on a metric space $X$. If the action is cocompact then every element of $\Gamma$ is a semisimple isometry of $X$.
(Please refer to Bridson and Haefliger's book, page 229. Here is a link.)
So I did this proof as follows and I never used the action is proper (ie. properly discontinuous). Can anyone suggest if there is a mistake? I use the notation from this book.
As the action is cocompact, $X=\cup \gamma_i K$ where $\gamma_i\in \Gamma$ and $K$ is a compact set in $X$. There is a sequence of points say $x_n$ such that $d_\gamma(x_n)$ converges to $|\gamma|$. So for each $x_n$ there is $\gamma_n$ such that $y_n=\gamma_n.x_n\in K$. As $K$ is compact there is a subsequence converging to say $y$. This $y$ gives us the minimal element we are looking for.
How do you justify that $y$ minimizes the displacement function $d_{\gamma}$? You know that $$d(\gamma \cdot y , y)= \lim\limits_{n \to + \infty} d(\gamma \gamma_n \cdot x_n, \gamma_n \cdot x_n)= \lim\limits_{n \to + \infty} d( \gamma_n^{-1} \gamma \gamma_n \cdot x_n,x_n), $$
but $\gamma_n^{-1} \gamma \gamma_n$ depends on $n$, and properness is precisely used in order to replace $\gamma_n^{-1} \gamma \gamma_n$ with some $\overline{\gamma}$ independent on $n$.
In fact, it is possible for a group $\Gamma$ to act cocompactly on a metric space $X$ with a parabolic isometry. Let $X= \mathbb{H}^2= \{ z \in \mathbb{C} \mid \mathrm{Im}(z)>0 \}$ be the hyperbolic plane and let us define the following two isometries:
$$h : z \mapsto \lambda \cdot z \ \ \ \mathrm{and} \ \ \ t : z \mapsto z+1,$$
where $\lambda >1$. It is not difficult to notice that $\Gamma= \langle h,t \rangle$ acts cocompactly on $X$ and that $t$ is a parabolic isometry. Of course, the action $\Gamma \curvearrowright X$ is not proper: for large enough $n$, $$d(h^{-n} t h^n(x),x)<1.$$