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 \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and \ref{spaces-morphisms-lemma-smooth-flat}). 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 \ref{lemma-property-spaces-too}. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} and Lemma \ref{spaces-morphisms-lemma-fppf-open}. This clearly implies that our definition is independent of the choice of $p : U \to \mathcal{X}$. \medskip\noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. By Algebraic Stacks, Lemma \ref{algebraic-lemma-lift-morphism-presentations} 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. \medskip\noindent 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 \ref{lemma-points-cartesian}. 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 \ref{spaces-morphisms-lemma-fppf-open}. 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.