Proof.
Choose a morphism p : U \to \mathcal{X} which is surjective, flat, and locally of finite presentation with U an algebraic space. Such exist by the definition of an algebraic stack, as a smooth morphism is flat and locally of finite presentation (see Morphisms of Spaces, Lemmas 67.37.5 and 67.37.7). We define a topology on |\mathcal{X}| by the rule: W \subset |\mathcal{X}| is open if and only if |p|^{-1}(W) is open in |U|. To show that this is independent of the choice of p, let p' : U' \to \mathcal{X} be another morphism which is surjective, flat, locally of finite presentation from an algebraic space to \mathcal{X}. Set U'' = U \times _\mathcal {X} U' so that we have a 2-commutative diagram
\xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & \mathcal{X} }
As U \to \mathcal{X} and U' \to \mathcal{X} are surjective, flat, locally of finite presentation we see that U'' \to U' and U'' \to U are surjective, flat and locally of finite presentation, see Lemma 100.3.2. Hence the maps |U''| \to |U'| and |U''| \to |U| are continuous, open and surjective, see Morphisms of Spaces, Definition 67.5.2 and Lemma 67.30.6. This clearly implies that our definition is independent of the choice of p : U \to \mathcal{X}.
Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks. By Algebraic Stacks, Lemma 94.15.1 we can find a 2-commutative diagram
\xymatrix{ U \ar[d]_ x \ar[r]_ a & V \ar[d]^ y \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} }
with surjective smooth vertical arrows. Consider the associated commutative diagram
\xymatrix{ |U| \ar[d]_{|x|} \ar[r]_{|a|} & |V| \ar[d]^{|y|} \\ |\mathcal{X}| \ar[r]^{|f|} & |\mathcal{Y}| }
of sets. If W \subset |\mathcal{Y}| is open, then by the definition above this means exactly that |y|^{-1}(W) is open in |V|. Since |a| is continuous we conclude that |a|^{-1}|y|^{-1}(W) = |x|^{-1}|f|^{-1}(W) is open in |W| which means by definition that |f|^{-1}(W) is open in |\mathcal{X}|. Thus |f| is continuous.
Finally, we have to show that if U is an algebraic space, and U \to \mathcal{X} is flat and locally of finite presentation, then |U| \to |\mathcal{X}| is open. Let V \to \mathcal{X} be surjective, flat, and locally of finite presentation with V an algebraic space. Consider the commutative diagram
\xymatrix{ |U \times _\mathcal {X} V| \ar[r]_ e \ar[rd]_ f & |U| \times _{|\mathcal{X}|} |V| \ar[d]_ c \ar[r]_ d & |V| \ar[d]^ b \\ & |U| \ar[r]^ a & |\mathcal{X}| }
Now the morphism U \times _\mathcal {X} V \to U is surjective, i.e, f : |U \times _\mathcal {X} V| \to |U| is surjective. The left top horizontal arrow is surjective, see Lemma 100.4.3. The morphism U \times _\mathcal {X} V \to V is flat and locally of finite presentation, hence d \circ e : |U \times _\mathcal {X} V| \to |V| is open, see Morphisms of Spaces, Lemma 67.30.6. Pick W \subset |U| open. The properties above imply that b^{-1}(a(W)) = (d \circ e)(f^{-1}(W)) is open, which by construction means that a(W) is open as desired.
\square
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