Proof.
The rough idea is to use that \mathcal{X} has a dense open which is a gerbe (Morphisms of Stacks, Proposition 101.29.1) and appeal to Lemma 106.10.2. The reason this does not work is that the open may not be quasi-compact and one runs into technical problems. Thus we first do a (standard) reduction to the Noetherian case.
First we choose a closed immersion \mathcal{X} \to \mathcal{X}' where \mathcal{X}' is an algebraic stack separated and of finite type over Y. See Limits of Stacks, Lemma 102.6.2. Clearly it suffices to prove the theorem for \mathcal{X}', hence we may assume \mathcal{X} \to Y is separated and of finite presentation.
Assume \mathcal{X} \to Y is separated and of finite presentation. By Limits of Spaces, Proposition 70.8.1 we can write Y = \mathop{\mathrm{lim}}\nolimits Y_ i as the directed limit of a system of Noetherian algebraic spaces with affine transition morphisms. By Limits of Stacks, Lemma 102.5.1 there is an i and a morphism \mathcal{X}_ i \to Y_ i of finite presentation from an algebraic stack to Y_ i such that \mathcal{X} = Y \times _{Y_ i} \mathcal{X}_ i. After increasing i we may assume that \mathcal{X}_ i \to Y_ i is separated, see Limits of Stacks, Lemma 102.4.2. Then it suffices to prove the theorem for \mathcal{X}_ i \to Y_ i. This reduces us to the case discussed in the next paragraph.
Assume Y is Noetherian. We may replace \mathcal{X} by its reduction (Properties of Stacks, Definition 100.10.4). This reduces us to the case discussed in the next paragraph.
Assume Y is Noetherian and \mathcal{X} is reduced. Since \mathcal{X} \to Y is separated and Y quasi-separated, we see that \mathcal{X} is quasi-separated as an algebraic stack. Hence the inertia \mathcal{I}_\mathcal {X} \to \mathcal{X} is quasi-compact. Thus by Morphisms of Stacks, Proposition 101.29.1 there exists a dense open substack \mathcal{V} \subset \mathcal{X} which is a gerbe. Let \mathcal{V} \to V be the morphism which expresses \mathcal{V} as a gerbe over the algebraic space V. See Morphisms of Stacks, Lemma 101.28.2 for a construction of \mathcal{V} \to V. This construction in particular shows that the morphism \mathcal{V} \to Y factors as \mathcal{V} \to V \to Y. Picture
\xymatrix{ \mathcal{V} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ V \ar[r] & Y }
Since the morphism \mathcal{V} \to V is surjective, flat, and of finite presentation (Morphisms of Stacks, Lemma 101.28.8) and since \mathcal{V} \to Y is locally of finite presentation, it follows that V \to Y is locally of finite presentation (Morphisms of Stacks, Lemma 101.27.12). Note that \mathcal{V} \to V is a universal homeomorphism (Morphisms of Stacks, Lemma 101.28.13). Since \mathcal{V} is quasi-compact (see Morphisms of Stacks, Lemma 101.8.2) we see that V is quasi-compact. Finally, since \mathcal{V} \to Y is separated the same is true for V \to Y by Morphisms of Stacks, Lemma 101.27.17 applied to \mathcal{V} \to V \to Y (whose assumptions are satisfied as we've already seen).
All of the above means that the assumptions of Limits of Spaces, Lemma 70.13.3 apply to the morphism V \to Y. Thus we can find a dense open subspace V' \subset V and an immersion V' \to \mathbf{P}^ n_ Y over Y. Clearly we may replace V by V' and \mathcal{V} by the inverse image of V' in \mathcal{V} (recall that |\mathcal{V}| = |V| as we've seen above). Thus we may assume we have a diagram
\xymatrix{ \mathcal{V} \ar[rr] \ar[d] & & \mathcal{X} \ar[d] \\ V \ar[r] & \mathbf{P}^ n_ Y \ar[r] & Y }
where the arrow V \to \mathbf{P}^ n_ Y is an immersion. Let \mathcal{X}' be the scheme theoretic image of the morphism
j : \mathcal{V} \longrightarrow \mathbf{P}^ n_ Y \times _ Y \mathcal{X}
and let Y' be the scheme theoretic image of the morphism V \to \mathbf{P}^ n_ Y. We obtain a commutative diagram
\xymatrix{ \mathcal{V} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & \mathbf{P}^ n_ Y \times _ Y \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ V \ar[r] & Y' \ar[r] & \mathbf{P}^ n_ Y \ar[r] & Y }
(See Morphisms of Stacks, Lemma 101.38.4). We claim that \mathcal{V} = V \times _{Y'} \mathcal{X}' and that Lemma 106.10.2 applies to the morphism \mathcal{X}' \to Y' and the open subspace V \subset Y'. If the claim is true, then we obtain
\xymatrix{ \overline{X} \ar[rd]_{\overline{g}} & X \ar[l] \ar[d]_ g \ar[r]_ h & \mathcal{X}' \ar[ld]^ f \\ & Y' }
with X \to \overline{X} an open immersion, \overline{g} and h proper, and such that |V| is contained in the image of |g|. Then the composition X \to \mathcal{X}' \to \mathcal{X} is proper (as a composition of proper morphisms) and its image contains |\mathcal{V}|, hence this composition is surjective. As well, \overline{X} \to Y' \to Y is proper as a composition of proper morphisms.
The last step is to prove the claim. Observe that \mathcal{X}' \to Y' is separated and of finite type, that Y' is quasi-compact and quasi-separated, and that V is quasi-compact (we omit checking all the details completely). Next, we observe that b : \mathcal{X}' \to \mathcal{X} is an isomorphism over \mathcal{V} by Morphisms of Stacks, Lemma 101.38.7. In particular \mathcal{V} is identified with an open substack of \mathcal{X}'. The morphism j is quasi-compact (source is quasi-compact and target is quasi-separated), so formation of the scheme theoretic image of j commutes with flat base change by Morphisms of Stacks, Lemma 101.38.5. In particular we see that V \times _{Y'} \mathcal{X}' is the scheme theoretic image of \mathcal{V} \to V \times _{Y'} \mathcal{X}'. However, by Morphisms of Stacks, Lemma 101.37.5 the image of |\mathcal{V}| \to |V \times _{Y'} \mathcal{X}'| is closed (use that \mathcal{V} \to V is a universal homeomorphism as we've seen above and hence is universally closed). Also the image is dense (combine what we just said with Morphisms of Stacks, Lemma 101.38.6) we conclude |\mathcal{V}| = |V \times _{Y'} \mathcal{X}'|. Thus \mathcal{V} \to V \times _{Y'} \mathcal{X}' is an isomorphism and the proof of the claim is complete.
\square
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