QUESTION: So I tried to see what ratcliffe meant here, but I don't get it. Namely, how does he get a contradiction at the end. I don't see how $P$ being locally finite gives us a contradiction. I assumed that Ratcliffe wanted a contradiction by noting that $g_i\overline{P}$ forms a covering of $B^n$ and by some finagling you get a disjoint infinite open cover of $K$, which contradicts compactness but that doesn't seem to work here.
EDIT: DEFINITION: A collection $\mathcal{S}$ of subsets of a topological space $X$ is locally finite if and only if for each point $x$ of $X$, there is an open neighborhood $U$ of $x$ in $X$ such that $U$ meets only finitely many members of $S$
DEFINITION: A convex fundamental polyhedron for a discrete group $\Gamma$ of isometries of $X$ is a convex polyhedron $P$ in $X$ whose interior is a locally finite fundamental domain for $\Gamma$.
DEFINITION: A subset $R$ of a metric space $X$ is a fudamental domain for agroup $\Gamma$ of isometries of $X$ iff
(1) The set $R$ is open in X;
(2) The members of ${gR:g\in\Gamma}$ are mutually disjoint; and
(3) $X=\cup{g\overline{R}:g\in\Gamma}$.
(4) $R$ is connected
EDIT2: Ok I think I see what the contradiction is. We note that $g_iP$ forms a locally finite collection for $K$. Let $x_j\in K$, then by local finiteness there exists an open subset $U_{x_j}$ that intersects with finitely many elements in our locally finite collection. By compactness we have a finite sub cover $U_{x_1},\cdots,U_{x_n}$. But this collection intersects with only finitely many elements in our infinite collection $g_iP$, and hence only finitely many interest $K$, which contradicts the fact that $K\cap g_iP\neq\emptyset$ for infinitely many elements.