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Choose $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ with ${\iota }^{\ast }(\hat{\zeta })=\zeta$, and letbe its extension. By the given fact, $E$ is a profinite group, hence is residually finite. The extension $E_{\zeta }^{\prime }$ over $\Gamma$ is the pullback of $E$ along the injective map $\iota$. Part 5(a)(iii) embeds $E_{\zeta }^{\prime }$ into $E$. Since every subgroup of a residually finite group is residually finite, so is $E_{\zeta }^{\prime }$.

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.