Suppose we have a commutative diagram $$ \xymatrix{ \mathcal{X}' \ar[d]_{f'} \ar[r] & \mathcal{X} \ar[d]^f \\ Y' \ar[r] & Y } $$ and a quasi-compact open $V' \subset Y'$, such that $Y' \to Y$ is a proper morphism of algebraic spaces, $\mathcal{X}' \to \mathcal{X}$ is a proper morphism of algebraic stacks, $V' \subset Y'$ maps surjectively onto $V$, and $\mathcal{X}'_{V'}$ is a gerbe over $V'$. Then it suffices to prove the lemma for the pair $(f' : \mathcal{X}' \to Y', V')$. Some details omitted. \medskip\noindent Overall strategy of the proof. We will reduce to the case where the image of $f$ is open and $f$ has a section over this open by repeatedly applying the above remark. Each step is straightforward, but there are quite a few of them which makes the proof a bit involved. \medskip\noindent Using Limits of Spaces, Proposition \ref{spaces-limits-proposition-there-is-a-scheme-finite-over} we reduce to the case where $Y$ is a scheme. (Let $Y' \to Y$ be a finite surjective morphism where $Y'$ is a scheme. Set $\mathcal{X}' = \mathcal{X}_{Y'}$ and apply the initial remark of the proof.) \medskip\noindent Using Lemma \ref{lemma-flatten-stack} (and Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-fppf} to see that a gerbe is flat and locally of finite presentation) we reduce to the case where $f$ is flat and of finite presentation. \medskip\noindent Since $f$ is flat and locally of finite presentation, we see that the image of $|f|$ is an open $W \subset Y$. Since $\mathcal{X}$ is quasi-compact (as $f$ is of finite type and $Y$ is quasi-compact) we see that $W$ is quasi-compact. By Lemma \ref{lemma-finite-cover-factor} we can find a finite surjective morphism $g : Y' \to Y$ such that $g^{-1}(W) \to Y$ factors Zariski locally through $\mathcal{X} \to Y$. After replacing $Y$ by $Y'$ and $\mathcal{X}$ by $\mathcal{X} \times_Y Y'$ we reduce to the situation described in the next paragraph. \medskip\noindent Assume there exists $n \geq 0$, quasi-compact opens $W_i \subset Y$, $i = 1, \ldots, n$, and morphisms $x_i : W_i \to \mathcal{X}$ such that (a) $f \circ x_i = \text{id}_{W_i}$, (b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, and (c) $W$ is the image of $|f|$. We will use induction on $n$. The base case is $n = 0$: this implies $V = \emptyset$ and in this case we can take $\overline{Z} = \emptyset$. If $n > 0$, then for $i = 1, \ldots, n$ consider the reduced closed subschemes $Y_i$ with underlying topological space $Y \setminus W_i$. Consider the finite morphism $$ Y' = Y \amalg \coprod\nolimits_{i = 1, \ldots, n} Y_i \longrightarrow Y $$ and the quasi-compact open $$ V' = (W_1 \cap \ldots \cap W_n \cap V) \amalg \coprod_{i = 1, \ldots, n} (V \cap Y_i). $$ By the initial remark of the proof, if we can prove the lemma for the pairs $$ (\mathcal{X} \to Y, W_1 \cap \ldots \cap W_n \cap V) \quad\text{and}\quad (\mathcal{X} \times_Y Y_i \to Y_i, V \cap Y_i),\quad i = 1, \ldots, n $$ then the result is true. Here we use the settheoretic equality $V = (W_1 \cap \ldots \cap W_n \cap V) \cup \bigcup\nolimits_{i = 1, \ldots n} (V \cap Y_i)$. The induction hypothesis applies to the second type of pairs above. Hence we reduce to the situation described in the next paragraph. \medskip\noindent Assume there exists $n \geq 0$, quasi-compact opens $W_i \subset Y$, $i = 1, \ldots, n$, and morphisms $x_i : W_i \to \mathcal{X}$ such that (a) $f \circ x_i = \text{id}_{W_i}$, (b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, (c) $W$ is the image of $|f|$, and (d) $V \subset W_1 \cap \ldots \cap W_n$. The morphisms $$ T_{ij} = \mathit{Isom}_\mathcal{X}(x_i|_{W_i \cap W_j \cap V}, x_j|_{W_i \cap W_j \cap V}) \longrightarrow W_i \cap W_j \cap V $$ are surjective, flat, and locally of finite presentation (Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-isom-fppf}). We apply Lemma \ref{lemma-finite-cover-factor} to each quasi-compact open $W_i \cap W_j \cap V$ and the morphisms $T_{ij} \to W_i \cap W_j \cap V$ to get finite surjective morphisms $Y'_{ij} \to Y$. After replacing $Y$ by the fibre product of all of the $Y'_{ij}$ over $Y$ we reduce to the situation described in the next paragraph. \medskip\noindent Assume there exists $n \geq 0$, quasi-compact opens $W_i \subset Y$, $i = 1, \ldots, n$, and morphisms $x_i : W_i \to \mathcal{X}$ such that (a) $f \circ x_i = \text{id}_{W_i}$, (b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_n$, and (e) $x_i$ and $x_j$ are Zariski locally isomorphic over $W_i \cap W_j \cap V$. Let $y \in V$ be arbitrary. Suppose that we can find a quasi-compact open neighbourhood $y \in V_y \subset V$ such that the lemma is true for the pair $(\mathcal{X} \to Y, V_y)$, say with solution $\overline{Z}_y, Z_y, \overline{g}_y, g_y, h_y$. Since $V$ is quasi-compact, we can find a finite number $y_1, \ldots, y_m$ such that $V = V_{y_1} \cup \ldots \cup V_{y_m}$. Then it follows that setting $$ \overline{Z} = \coprod \overline{Z}_{y_j},\quad Z = \coprod Z_{y_j},\quad \overline{g} = \coprod \overline{g}_{y_j},\quad g = \coprod g_{y_j},\quad h = \coprod h_{y_j} $$ is a solution for the lemma. Given $y$ by condition (e) we can choose a quasi-compact open neighbourhood $y \in V_y \subset V$ and isomorphisms $\varphi_i : x_1|_{V_y} \to x_i|_{V_y}$ for $i = 2, \ldots, n$. Set $\varphi_{ij} = \varphi_j \circ \varphi_i^{-1}$. This leads us to the situation described in the next paragraph. \medskip\noindent Assume there exists $n \geq 0$, quasi-compact opens $W_i \subset Y$, $i = 1, \ldots, n$, and morphisms $x_i : W_i \to \mathcal{X}$ such that (a) $f \circ x_i = \text{id}_{W_i}$, (b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_n$, and (f) there are isomorphisms $\varphi_{ij} : x_i|_V \to x_j|_V$ satisfying $\varphi_{jk} \circ \varphi_{ij} = \varphi_{ik}$. The morphisms $$ I_{ij} = \mathit{Isom}_\mathcal{X}(x_i|_{W_i \cap W_j}, x_j|_{W_i \cap W_j}) \longrightarrow W_i \cap W_j $$ are proper because $f$ is separated (Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-separated-implies-isom}). Observe that $\varphi_{ij}$ defines a section $V \to I_{ij}$ of $I_{ij} \to W_i \cap W_j$ over $V$. By More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-get-section-after-blowup} we can find $V$-admissible blowups $p_{ij} : Y_{ij} \to Y$ such that $s_{ij}$ extends to $p_{ij}^{-1}(W_i \cap W_j)$. After replacing $Y$ by the fibre product of all the $Y_{ij}$ over $Y$ we get to the situation described in the next paragraph. \medskip\noindent Assume there exists $n \geq 0$, quasi-compact opens $W_i \subset Y$, $i = 1, \ldots, n$, and morphisms $x_i : W_i \to \mathcal{X}$ such that (a) $f \circ x_i = \text{id}_{W_i}$, (b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_n$, and (g) there are isomorphisms $\varphi_{ij} : x_i|_{W_i \cap W_j} \to x_j|_{W_i \cap W_j}$ satisfying $$ \varphi_{jk}|_V \circ \varphi_{ij}|_V = \varphi_{ik}|_V. $$ After replacing $Y$ by another $V$-admissible blowup if necessary we may assume that $V$ is dense and scheme theoretically dense in $Y$ and hence in any open subspace of $Y$ containing $V$. After such a replacement we conclude that $$ \varphi_{jk}|_{W_i \cap W_j \cap W_k} \circ \varphi_{ij}|_{W_i \cap W_j \cap W_k} = \varphi_{ik}|_{W_i \cap W_j \cap W_k} $$ by appealing to Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-equality-of-morphisms} and the fact that $I_{ik} \to W_i \cap W_j$ is proper (hence separated). Of course this means that $(x_i, \varphi_{ij})$ is a desent datum and we obtain a morphism $x : W \to \mathcal{X}$ agreeing with $x_i$ over $W_i$ because $\mathcal{X}$ is a stack. Since $x$ is a section of the separated morphism $\mathcal{X} \to W$ we see that $x$ is proper (Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-section-immersion}). Thus the lemma now holds with $\overline{Z} = Y$, $Z = W$, $\overline{g} = \text{id}_Y$, $g = \text{id}_W$, $h = x$.