An acyclic noncontractible cone

The displayed source complex is acyclic: multiplication by $2$ is injective, reduction modulo $2$ is surjective, and its kernel is $2\mathbb{Z}$, the image of the first map. If it were contractible, [L1] would make its identity null-homotopic; in degree $0$ that would supply a section $\mathbb{Z}\mathrm{/}2\to \mathbb{Z}$ of the quotient map $\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, which is impossible. Thus the source complex is acyclic and noncontractible.

By [L2], the cone of the displayed zero map is the shift of the source complex. Because the zero map from an acyclic complex to the zero complex induces isomorphisms on all homology groups, [L4] makes that map a quasi-isomorphism; then [L3] makes its shift a quasi-isomorphism too. Hence the cone is acyclic. If the cone were contractible, applying [L3] with shift $[-1]$ would make the source complex contractible, contradicting step 1.1. Thus this cone is acyclic and noncontractible, so it refutes the displayed statement.