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The local isometry from a connected special cube complex to a Salvetti complex induces an injection of fundamental groups. Hence its fundamental group is a subgroup of a Right-angled Artin group.

For a finite graph $\Gamma$, its Right-angled Artin group isIts Salvetti complex $S_{\Gamma }$ is the one-vertex cube complex with an oriented loop labelled $v$ for each vertex of $\Gamma$, a square torus for every edge, and more generally one cubulated $k$-torus for every $k$-clique, attached compatibly along coordinate subtori. Thus ${\pi }_1(S_{\Gamma })\cong A_{\Gamma }$. The vertex link of a Salvetti complex is flag, so $S_{\Gamma }$ is a nonpositively curved cube complex.

Because $X$ is special, the fundamental group of a special cube complex embeds in a Right-angled Artin group. Right-angled Artin groups are residually finite by the residual finiteness of a right-angled Artin group, and a subgroup of a residually finite group is residually finite. Hence ${\pi }_{1}X$ is residually finite.

If ${\pi }_{1}X$ were simple, choose $1\ne g\in {\pi }_{1}X$. A finite quotient in which $g$ survives has a proper normal kernel. Simplicity would force that kernel to be trivial, embedding ${\pi }_{1}X$ into a finite group, contrary to the assumption that ${\pi }_{1}X$ is infinite. This is precisely the obstruction that an infinite residually finite group is not simple.