 Solution

Retain the metric geodesic subdivision and orbit-density parts of the usual Milnor–Švarc lemma proof. Drop the step using proper discontinuity and compactness to prove that the displacement-bounded generating set is finite.

More explicitly, let $L=d_X(x,gx)$ and $m=\max \{1,\lceil L\rceil \}$. Subdivide a metric geodesic from $x$ to $gx$ into $m$ segments of length at most one. Choose orbit points $g_{i}x$ within distance $R$ of the subdivision points, choosing the endpoints exactly: $g_0=1$, $g_m=g$. Thenso every nonidentity increment $g_{i-1}^{-1}{g}_i$ lies in $S$. Their telescoping product is $g$, and henceThis also covers $L=0$, when the one increment may be a nontrivial stabilizer element. Conversely, for any word $g=s_1\cdots s_k$ in $S$, the triangle inequality and invariance under the isometric action giveTaking the shortest word gives the other inequality. Thus the proof works unchanged except that the resulting generating set may be infinite. Neither local compactness nor a finite stabilizer is used in this weaker argument.