A warmup for this proof is the proof of Lemma \ref{lemma-basis-hypercovering} and we encourage the reader to read that proof first. \medskip\noindent First we replace $\mathcal{C}$ by the site $\mathcal{C}/X$. After doing so we may assume that $X$ is the final object of $\mathcal{C}$ and that $\mathcal{C}$ has all finite limits (Categories, Lemma \ref{categories-lemma-finite-limits-exist}). \medskip\noindent Let $n \geq 0$. Let us say that an {\it $n$-truncated $\mathcal{B}$-hypercovering of $X$} is given by an $n$-truncated simplicial object $K$ of $\text{SR}(\mathcal{C})$ such that for $i \in I_a$, $0 \leq a \leq n$ we have $U_{a, i} \in \mathcal{B}$ and such that $K_0$ is a covering of $X$ and $K_{a + 1} \to (\text{cosk}_a \text{sk}_a K)_{a + 1}$ for $a = 0, \ldots, n - 1$ is a covering as in Definition \ref{definition-covering-SR}. \medskip\noindent Since $X$ has a covering $\{U_{0, i} \to X\}_{i \in I_0}$ with $U_i \in \mathcal{B}$ by assumption, we get a $0$-truncated $\mathcal{B}$-hypercovering of $X$. Observe that any $0$-truncated $\mathcal{B}$-hypercovering of $X$ is split, see Lemma \ref{lemma-split}. \medskip\noindent The lemma follows if we can prove for $n \geq 0$ that given a split $n$-truncated $\mathcal{B}$-hypercovering $K$ of $X$ we can extend it to a split $(n + 1)$-truncated $\mathcal{B}$-hypercovering of $X$. \medskip\noindent Construction of the extension. Consider the $(n + 1)$-truncated simplicial object $K' = \text{sk}_{n + 1}(\text{cosk}_n K)$ of $\text{SR}(\mathcal{C})$. Write $$ K'_{n + 1} = \{U'_{n + 1, i}\}_{i \in I'_{n + 1}} $$ Since $K = \text{sk}_n K'$ we have $K_a = K'_a$ for $0 \leq a \leq n$. For every $i' \in I'_{n + 1}$ we choose a covering \begin{equation} \label{equation-choose-covering-B} \{g_{n + 1, j} : U_{n + 1, j} \to U'_{n + 1, i'}\}_{j \in J_{i'}} \end{equation} with $U_{n + 1, j} \in \mathcal{B}$ for $j \in J_{i'}$. This is possible by our assumption on $\mathcal{B}$ in the lemma. For $0 \leq m \leq n$ denote $N_m \subset I_m$ the subset of nondegenerate indices. We set $$ I_{n + 1} = \coprod\nolimits_{\varphi : [n + 1] \to [m]\text{ surjective, }0\leq m \leq n} N_m \amalg \coprod\nolimits_{i' \in I'_{n + 1}} J_{i'} $$ For $j \in I_{n + 1}$ we set $$ U_{n + 1, j} = \left\{ \begin{matrix} U_{m, i} & \text{if} & j = (\varphi, i) & \text{where} & \varphi : [n + 1] \to [m], i \in N_m \\ U_{n + 1, j} & \text{if} & j \in J_{i'} & \text{where} & i' \in I'_{n + 1} \end{matrix} \right. $$ with obvious notation. We set $K_{n + 1} = \{U_{n + 1, j}\}_{j \in I_{n + 1}}$. By construction $U_{n + 1, j}$ is an element of $\mathcal{B}$ for all $j \in I_{n + 1}$. Let us define compatible maps $$ I_{n + 1} \to I'_{n + 1} \quad\text{and}\quad K_{n + 1} \to K'_{n + 1} $$ Namely, the first map is given by $(\varphi, i) \mapsto \alpha'(\varphi)(i)$ and $(j \in J_{i'}) \mapsto i'$. For the second map we use the morphisms $$ f'_{\varphi, i} : U_{m, i} \to U'_{n + 1, \alpha'(\varphi)(i)} \quad\text{and}\quad g_{n + 1, j} : U_{n + 1, j} \to U'_{n + 1, i'} $$ We claim the morphism $$ K_{n + 1} \to K'_{n + 1} = (\text{cosk}_n \text{sk}_n K')_{n + 1} = (\text{cosk}_n K)_{n + 1} $$ is a covering as in Definition \ref{definition-covering-SR}. Namely, if $i' \in I'_{n + 1}$, then either $i'$ is nondegenerate and the inverse image of $i'$ in $I_{n + 1}$ is equal to $J_{i'}$ and we get a covering of $U'_{n + 1, i'}$ by our choice (\ref{equation-choose-covering-B}), or $i'$ is degenerate and the inverse image of $i'$ in $I_{n + 1}$ is $J_{i'} \amalg \{(\varphi, i)\}$ for a unique pair $(\varphi, i)$ and we get a covering by our choice (\ref{equation-choose-covering-B}) and assumption (2) of the lemma. \medskip\noindent To finish the proof we have to define the morphisms $K(\varphi) : K_{n + 1} \to K_m$ corresponding to morphisms $\varphi : [m] \to [n + 1]$, $0 \leq m \leq n$ and the morphisms $K(\varphi) : K_m \to K_{n + 1}$ corresponding to morphisms $\varphi : [n + 1] \to [m]$, $0 \leq m \leq n$ satisfying suitable composition relations. For the first kind we use the composition $$ K_{n + 1} \to K'_{n + 1} \xrightarrow{K'(\varphi)} K'_m = K_m $$ to define $K(\varphi) : K_{n + 1} \to K_m$. For the second kind, suppose given $\varphi : [n + 1] \to [m]$, $0 \leq m \leq n$. We define the corresponding morphism $K(\varphi) : K_m \to K_{n + 1}$ as follows: \begin{enumerate} \item for $i \in I_m$ there is a unique surjective map $\psi : [m] \to [m_0]$ and a unique $i_0 \in I_{m_0}$ nondegenerate such that $\alpha(\psi)(i_0) = i$\footnote{For example, if $i$ is nondegenerate, then $m = m_0$ and $\psi = \text{id}_{[m]}$.}, \item we set $\varphi_0 = \psi_0 \circ \varphi : [n + 1] \to [m_0]$ and we map $i \in I_m$ to $(\varphi_0, i_0) \in I_{n + 1}$, in other words, $\alpha(\varphi)(i) = (\varphi_0, i_0)$, and \item the morphism $f_{\varphi, i} : U_{m, i} \to U_{n + 1, \alpha(\varphi)(i)} = U_{m_0, i_0}$ is the inverse of the isomorphism $f_{\psi, i_0} : U_{m_0, i_0} \to U_{m, i}$ (see Lemma \ref{lemma-split}). \end{enumerate} We omit the straightforward but cumbersome verification that this defines a split $(n + 1)$-truncated $\mathcal{B}$-hypercovering of $X$ extending the given $n$-truncated one. In fact, everything is clear from the above, except for the verification that the morphisms $K(\varphi)$ compose correctly for all $\varphi : [a] \to [b]$ with $0 \leq a, b \leq n + 1$.