Proof (due to Joseph Bernstein). Assume that $H^i(X^\bullet)=0$ for $i>n$. We choose a $G$-morphism $A^n\to \ker[X^n\to X^{n+1}]$ such that the induced morphism $A^n\to H^n(X^\bullet)$ is surjective, where $A^n$ is a finitely generated (over $\mathbb{Z}$) torsion-free $G$-module. We regard $A^n$ as a complex (with one $G$-module $A^n$ in degree $n$). We have a morphism of complexes $\varphi\colon A^n\to X^\bullet$. We denote by $X_{(1)}^\bullet$ the cone of $\varphi$. It is easy to see that $H^n(X_{(1)}^\bullet)=0$. Then we apply this procedure to $X_{(1)}^\bullet$ for $n-1$ to obtain $X_{(2)}^\bullet$ with $H^{n-1}(X_{(2)}^\bullet)=0$, and so on.

Assume that $H^i(X^\bullet)=0$ for $i\le n-m$. Then the complex $X_{(m)}^\bullet$ is acyclic. One can check that $X_{(m)}^\bullet$ is the cone of some morphism of complexes $\psi\colon M^\bullet\to X^\bullet$, where $M^\bullet$ is a bounded complex of finitely generated torsion-free $G$-modules. Since the cone $X_{(m)}^\bullet$ of $\psi$ is acyclic, we see that $\psi$ is a quasi-isomorphism.