**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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