When ${\mathcal{X}}$ is snc, the result is a direct consequence of (i) and (ii) above. When ${\mathcal{X}}$ is dlt, we have by definition

$$\operatorname{Sk}({\mathcal{X}})=\operatorname{Sk}({\mathcal{X}}_{\mathrm{snc}})\subset{\mathcal{X}}_{\mathrm{snc}}^{\mathrm{an}}\subset{\mathcal{X}}^{\mathrm{an}},$$

and $A_{\mathcal{X}}=A_{{\mathcal{X}}_{\mathrm{snc}}}$ on ${\mathcal{X}}_{\mathrm{snc}}^{\mathrm{an}}$. It is thus enough to show that any $v\in{\mathcal{X}}^{\mathrm{an}}$ with $A_{\mathcal{X}}(v)=0$ belongs to ${\mathcal{X}}_{\mathrm{snc}}^{\mathrm{an}}$, i.e. satisfies $c_{\mathcal{X}}(v)\in{\mathcal{X}}_{\mathrm{snc}}$. But $c_{\mathcal{X}}(v)$ is an lc center by Lemma 5.7, and hence $c_{\mathcal{X}}(v)\in{\mathcal{X}}_{\mathrm{snc}}$ by definition of dlt singularities. ∎