In this proof we write $\mathcal{C} = \mathcal{X}_\etale$ (resp.\ $\mathcal{C} = \mathcal{X}_{fppf}$) and we denote $\mathcal{C}' = \mathcal{X}_{lisse,\etale}$ (resp.\ $\mathcal{C}' = \mathcal{X}_{flat, fppf}$). Then $\mathcal{C}'$ is a full subcategory of $\mathcal{C}$. In this proof we will think of objects $V$ of $\mathcal{C}$ as schemes over $\mathcal{X}$ and objects $U$ of $\mathcal{C}'$ as schemes smooth (resp.\ flat) over $\mathcal{X}$. Finally, we write $\mathcal{O} = \mathcal{O}_\mathcal{X}$ and $\mathcal{O}' = \mathcal{O}_{\mathcal{X}_{lisse,\etale}}$ (resp.\ $\mathcal{O}' = \mathcal{O}_{\mathcal{X}_{flat,fppf}}$). In the notation above we have $\mathcal{O}(V) = \Gamma(V, \mathcal{O}_V)$ and $\mathcal{O}'(U) = \Gamma(U, \mathcal{O}_U)$. Consider the $\mathcal{O}$-module homomorphism $g_!\mathcal{O}' \to \mathcal{O}$ adjoint to the identification $\mathcal{O}' = g^{-1}\mathcal{O}$. \medskip\noindent Recall that $g_!\mathcal{O}'$ is the sheaf associated to the presheaf $g_{p!}\mathcal{O}'$ given by the rule $$ V \longmapsto \colim_{V \to U} \mathcal{O}'(U) $$ where the colimit is taken in the category of abelian groups (Modules on Sites, Definition \ref{sites-modules-definition-g-shriek}). Below we will use frequently that if $$ V \to U \to U' $$ are morphisms and if $f' \in \mathcal{O}'(U')$ restricts to $f \in \mathcal{O}'(U)$, then $(V \to U, f)$ and $(V \to U', f')$ define the same element of the colimit. Also, $g_!\mathcal{O}' \to \mathcal{O}$ maps the element $(V \to U, f)$ simply to the pullback of $f$ to $V$. \medskip\noindent To see that $g_!\mathcal{O}' \to \mathcal{O}$ is surjective it suffices to show that $1 \in \Gamma(\mathcal{C}, \mathcal{O})$ is locally in the image. Choose an object $U$ of $\mathcal{C}'$ corresponding to a surjective smooth morphism $U \to \mathcal{X}$. Then viewing $U$ both as an object of $\mathcal{C}'$ and $\mathcal{C}$ we see that $(U \to U, 1)$ is an element of the colimit above which maps to $1 \in \mathcal{O}(U)$. Since $U$ surjects onto the final object of $\Sh(\mathcal{C})$ we conclude $g_!\mathcal{O}' \to \mathcal{O}$ is surjective. \medskip\noindent Suppose that $s \in g_!\mathcal{O}'(V)$ is a section mapping to zero in $\mathcal{O}(V)$. To finish the proof we have to show that $s$ is zero. After replacing $V$ by the members of a covering we may assume $s$ is an element of the colimit $$ \colim_{V \to U} \mathcal{O}'(U) $$ Say $s = \sum (\varphi_i, s_i)$ is a finite sum with $\varphi_i : V \to U_i$, $U_i$ smooth (resp.\ flat) over $\mathcal{X}$, and $s_i \in \Gamma(U_i, \mathcal{O}_{U_i})$. Choose a scheme $W$ surjective \'etale over the algebraic space $U = U_1 \times_\mathcal{X} \ldots \times_\mathcal{X} U_n$. Note that $W$ is still smooth (resp.\ flat) over $\mathcal{X}$, i.e., defines an object of $\mathcal{C}'$. The fibre product $$ V' = V \times_{(\varphi_1, \ldots, \varphi_n), U} W $$ is surjective \'etale over $V$, hence it suffices to show that $s$ maps to zero in $g_!\mathcal{O}'(V')$. Note that the restriction $\sum (\varphi_i, s_i)|_{V'}$ corresponds to the sum of the pullbacks of the functions $s_i$ to $W$. In other words, we have reduced to the case of $(\varphi, s)$ where $\varphi : V \to U$ is a morphism with $U$ in $\mathcal{C}'$ and $s \in \mathcal{O}'(U)$ restricts to zero in $\mathcal{O}(V)$. By the commutative diagram $$ \xymatrix{ V \ar[rr]_-{(\varphi, 0)} \ar[rrd]_\varphi & & U \times \mathbf{A}^1 \\ & & U \ar[u]_{(\text{id}, 0)} } $$ we see that $((\varphi, 0) : V \to U \times \mathbf{A}^1, \text{pr}_2^*x)$ represents zero in the colimit above. Hence we may replace $U$ by $U \times \mathbf{A}^1$, $\varphi$ by $(\varphi, 0)$ and $s$ by $\text{pr}_1^*s + \text{pr}_2^*x$. Thus we may assume that the vanishing locus $Z : s = 0$ in $U$ of $s$ is smooth (resp.\ flat) over $\mathcal{X}$. Then we see that $(V \to Z, 0)$ and $(\varphi, s)$ have the same value in the colimit, i.e., we see that the element $s$ is zero as desired.