By Proposition 3.3.2, it is enough to show that $\mathrm{Sk}(\mathscr{X})$ is contained in the right hand side of (3.3.5). Shrinking $\mathscr{C}$ around $s$ if necessary, we can assume that $mK_{\mathscr{X}}+m(\mathscr{X}_{s})_{\mathrm{red}}$ is generated by global sections $\omega_{1},\ldots,\omega_{r}$. Then for each point $x$ on $\mathscr{X}^{\mathrm{snc}}_{s}$, we can choose an index $i$ in $\{1,\ldots,r\}$ such that $\mathrm{div}_{\mathscr{X}^{\mathrm{snc}}}(\omega_{i})+m(\mathscr{X}_{s})_{\mathrm{red}}$ is an effective divisor on $\mathscr{X}$ and $x$ is not contained in its support. This implies that the weight $\mathrm{wt}_{\omega_{i}}$ of $\omega_{i}$ is zero at all points of $\mathrm{Sk}(\mathscr{X})\cap\mathrm{red}_{\mathscr{X}}^{-1}(x)$ and non-negative at all other points of $X^{\mathrm{an}}_{K}$. Thus $\mathrm{Sk}(\mathscr{X})\cap\mathrm{red}_{\mathscr{X}}^{-1}(x)$ is contained in $\mathrm{Sk}(X_{K},\omega_{i})$. Varying the point $x$, we find that $\mathrm{Sk}(\mathscr{X})$ is contained in

$$\bigcup_{i=1}^{r}\mathrm{Sk}(X_{K},\omega_{i}).$$

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