Milnor–Švarc lemma

Fix $x_0 \in X$ and, using compactness of $X/G$, choose $R$ large enough that the $G$-translates of the closed ball $\bar B(x_0,R)$ cover $X$. Let $S = \{\, g \in G : g \ne e,\ d_X(x_0,gx_0) \le 2R+1 \,\}$, a finite set by proper discontinuity. To see $S$ generates $G$: given $g \in G$, mark points $x_0 = g_0 x_0, g_1 x_0, \dots, g_n x_0 = g x_0$ spaced at most $2R$ apart along a geodesic from $x_0$ to $g x_0$; each consecutive pair satisfies $d_X(g_{i-1}x_0, g_i x_0) \le 2R$, so $g_{i-1}^{-1} g_i \in S$, and multiplying these $n \le d_X(x_0, g x_0)/(2R) + 1$ elements of $S$ recovers $g$. Hence $d_S(e,g) \le C_1\, d_X(x_0,gx_0) + C_1$ for a constant $C_1$ depending only on $R$. Conversely, each generator moves $x_0$ by at most $2R+1$, so $d_X(x_0,gx_0) \le (2R+1)\, d_S(e,g)$. These two inequalities show $g \mapsto g x_0$ is a $(\lambda,\varepsilon)$-quasi-isometric embedding for suitable $\lambda,\varepsilon$, and cocompactness (every point of $X$ lies within $R$ of some $G$-translate of $x_0$) makes its image coarsely dense — so it is a quasi-isometry, and since it is defined on all of $G$, the finite set $S$ generates $G$.