Finite models for torsion-free lattices

I'll try to summarize YCor's comments into an answer (using big guns): Let $G$ be the real points of an algebraic group (a restriction by the OP in the comments) and assume $\Gamma$ irreducible.

Then Raghunathan shows that the answer is "yes" if $\Gamma$ is arithmetic. Margulis (Discrete subgroups of semisimple Lie groups) says that $\Gamma$ will be arithmetic if $G$ has rank at least $2$ (this is the sum over the ranks of the almost-factors). This leaves the case where $G$ is a single rank-1 factor. In that case $\Gamma\backslash G/K$ is a finite volume hyperbolic manifold from which one can cut off the cusps, see Theorem 12.7.2 in Ratcliffe, "Foundations of hyperbolic manifolds".

Edit: I'll also try to summarize the discussion about the reduction to the irreducible case. Suppose $G = G_1 \times \ldots \times G_k$ and the image of $\Gamma$ under the projection to $G_i$ is an irreducible lattice $\Gamma_i$. Let $\Gamma'' = \Gamma_1 \times \ldots \times \Gamma_k$. Now the above discussion shows that each $\Gamma_i$ acts properly and cocompactly on a contractible CW-complex $X_i$. Hence $\Gamma''$ acts properly and cocompactly on $X = X_1 \times \ldots \times X_k$. This action restricts to a proper and cocompact action of $\Gamma$. If $\Gamma$ happens to be torsion-free, the action is free and $\Gamma \backslash X$ is a finite model for $B\Gamma$.


In fact, more is true and you do not need separate arguments for rank 1 and higher rank.

The following is Theorem 13.1(i) in the book of Ballmann, Gromov and Schroeder "Manifolds of nonpositive curvature":

Suppose that $(M,g)$ is a complete real-analytic Riemannian manifold of nonpositive curvature and finite volume. Then $M$ is tame: It is diffeomorphic to the interior of a compact manifold with boundary $M'$.

Moreover, the proof shows that $M'$ can be realized as a submanifold (with boundary) of $M$.

Applying this to the locally-symmetric space $(M,g)=X/\Gamma$, where $\Gamma$ is a torsion-free lattice in the isometry group of a nonpositively curved symmetric space $X$, after triangulating $M'$, we obtain a finite model for $\Gamma$.