Proof.
(All of the morphisms occurring in this paragraph are representable by algebraic spaces, hence the conventions and results of Properties of Stacks, Section 100.3 are applicable.) Assume x is a finite type point. Choose an affine scheme U, a closed point u \in U, and a smooth morphism \varphi : U \to \mathcal{X} with \varphi (u) = x, see Lemma 101.18.3. Set u = \mathop{\mathrm{Spec}}(\kappa (u)) as usual. Set R = u \times _\mathcal {X} u so that we obtain a groupoid in algebraic spaces (u, R, s, t, c), see Algebraic Stacks, Lemma 94.16.1. The projection morphisms R \to u are the compositions
R = u \times _\mathcal {X} u \to u \times _\mathcal {X} U \to u \times _\mathcal {X} X = u
where the first arrow is of finite type (a base change of the closed immersion of schemes u \to U) and the second arrow is smooth (a base change of the smooth morphism U \to \mathcal{X}). Hence s, t : R \to u are locally of finite type (as compositions, see Morphisms of Spaces, Lemma 67.23.2). Since u is the spectrum of a field, it follows that s, t are flat and locally of finite presentation (by Morphisms of Spaces, Lemma 67.28.7). We see that \mathcal{Z} = [u/R] is an algebraic stack by Criteria for Representability, Theorem 97.17.2. By Algebraic Stacks, Lemma 94.16.1 we obtain a canonical morphism
f : \mathcal{Z} \longrightarrow \mathcal{X}
which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma 94.15.2 and a monomorphism, see Properties of Stacks, Lemma 100.8.4. It follows that the residual gerbe \mathcal{Z}_ x \subset \mathcal{X} of \mathcal{X} at x exists and that f factors through an equivalence \mathcal{Z} \to \mathcal{Z}_ x, see Properties of Stacks, Lemma 100.11.12. By construction the diagram
\xymatrix{ u \ar[d] \ar[r] & U \ar[d] \\ \mathcal{Z} \ar[r]^ f & \mathcal{X} }
is commutative. By Criteria for Representability, Lemma 97.17.1 the left vertical arrow is surjective, flat, and locally of finite presentation. Consider
\xymatrix{ u \times _\mathcal {X} U \ar[d] \ar[r] & \mathcal{Z} \times _\mathcal {X} U \ar[r] \ar[d] & U \ar[d] \\ u \ar[r] & \mathcal{Z} \ar[r]^ f & \mathcal{X} }
As u \to \mathcal{X} is locally of finite type, we see that the base change u \times _\mathcal {X} U \to U is locally of finite type. Moreover, u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U is surjective, flat, and locally of finite presentation as a base change of u \to \mathcal{Z}. Thus \{ u \times _\mathcal {X} U \to \mathcal{Z} \times _\mathcal {X} U\} is an fppf covering of algebraic spaces, and we conclude that \mathcal{Z} \times _\mathcal {X} U \to U is locally of finite type by Descent on Spaces, Lemma 74.16.1. By definition this means that f is locally of finite type (because the vertical arrow \mathcal{Z} \times _\mathcal {X} U \to \mathcal{Z} is smooth as a base change of U \to \mathcal{X} and surjective as \mathcal{Z} has only one point). Since \mathcal{Z} = \mathcal{Z}_ x we see that (3) holds.
It is clear that (3) implies (2). If (2) holds then x is a finite type point of \mathcal{X} by Lemma 101.18.4 and Lemma 101.18.6 to see that \mathcal{Z}_{\text{ft-pts}} is nonempty, i.e., the unique point of \mathcal{Z} is a finite type point of \mathcal{Z}.
\square
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