Proper and cocompact group action by isometries on a metric space

I don't think there is any missing assumption. Indeed cocompactness is enough to take advantage of compactness.

Let me also clarify the properness assumption. A action of a discrete group on a metric space is topologically proper (resp. metrically proper) if for every compact (resp. bounded) subset $K$, the set of $g\in G$ such that $K\cap gK$ is finite. "Metrically proper" implies "topologically proper", and the book uses "proper" to mean "metrically proper".

The definition of proper in the Bridson-Haefliger book should be taken as "metrically proper". (Their definition is flawed and should not be used as the given result fails otherwise, see the comments.)

Write $h_n=\gamma_n\gamma\gamma_n^{-1}$. They justify that $d(h_ny_n,y_n)$ is bounded. It follows that $d(h_nx,x)$ is bounded for every $x$ (namely by $2\sup_n d(x,y_n)$, which is bounded since $y_n\in K$). Metric properness of the action precisely means that every sequence $(h_n)$ such that $d(h_nx,x)$ is bounded for some $x$, is bounded (i.e., lies in one finite subset).

Not only properness of the space is not a missing assumption, but it is a consequence of the assumptions.

Proposition. Let $X$ be a metric space with a cocompact metrically proper action of a discrete group $G$. Then $X$ is a proper metric space.

[The action is not supposed isometric, not even continuous. For an isometric action properness is equivalent to $d(gx,x)\to\infty$ when $g\in\infty$, for some $x\in X$ and hence every $x$.]

Proof. Let $(x_n)$ be a bounded sequence in $X$. Write $x_n=g_nk_n$ with $g_n\in G$, $k_n\in K$. By metric properness of the action, $g_n$ remains in a finite subset. So, after extracting, we can suppose $g_n$ constant, say $g_n=g$, so $x_n=gk_n$. Hence for any limit point $k$ of $(k_n)$, $gk$ is a limit point.