Proof.
To prove that f is separated we have to show that \Delta : \mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} is proper. We already know that \Delta is representable by algebraic spaces, locally of finite type (Morphisms of Stacks, Lemma 101.3.3) and quasi-compact and quasi-separated (by definition of f being quasi-separated). Choose a scheme U and a surjective smooth morphism U \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}. Set
V = \mathcal{X} \times _{\Delta , \mathcal{X} \times _\mathcal {Y} \mathcal{X}} U
It suffices to show that the morphism of algebraic spaces V \to U is proper (Properties of Stacks, Lemma 100.3.3). Observe that U is locally Noetherian (use Morphisms of Stacks, Lemma 101.17.5 and the fact that U \to \mathcal{Y} is locally of finite type) and V \to U is of finite type and quasi-separated (as the base change of \Delta and properties of \Delta listed above). Applying Cohomology of Spaces, Lemma 69.19.2 it suffices to show: Given a commutative diagram
\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r]_ v \ar[d]_ j & V \ar[d]^ g \ar[r] & \mathcal{X} \ar[d]^\Delta \\ \mathop{\mathrm{Spec}}(A) \ar[r]^ u \ar@{-->}[ru] \ar@{..>}[rru] & U \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} }
where A is a discrete valuation ring and K its fraction field, there is a unique dashed arrow making the diagram commute. By Morphisms of Stacks, Lemma 101.39.4 the categories of dashed and dotted arrows are equivalent. Assumption (3) implies there is a unique dotted arrow up to isomorphism, see Morphisms of Stacks, Lemma 101.41.1. We conclude there is a unique dashed arrow as desired.
\square
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