Proof.
If (2) holds, then X_{red} is a separated algebraic space by applying Properties of Spaces, Lemma 66.3.3 to morphisms U \to X_{red} and V \to X_{red} with U, V affine and using that U \times _{X_{red}} V = U \times _ X V.
Assume (1). Let U \to X and V \to X be as in (2). Observe that U \times _ X V is a scheme by Lemma 87.11.2. Let U_{red}, V_{red}, X_{red} be the reduction of U, V, X. Then
U_{red} \times _{X_{red}} V_{red} = U_{red} \times _ X V_{red} \to U \times _ X V
is a thickening of schemes. It follows that (U \times _ X V)_{red} = (U_{red} \times _{X_{red}} V_{red})_{red}. In particular, we see that (U \times _ X V)_{red} is an affine scheme and that
\mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}((U \times _ X V)_{red})
is surjective, see Properties of Spaces, Lemma 66.3.3. Then U \times _ X V is affine by Limits of Spaces, Proposition 70.15.2. On the other hand, the morphism U \times _ X V \to U \times V of affine schemes is the composition
U \times _ X V = X \times _{(X \times _ S X)} (U \times _ S V) \to U \times _ S V \to U \times V
The first morphism is a monomorphism and locally of finite type (Lemma 87.11.2). The second morphism is an immersion (Schemes, Lemma 26.21.9). Hence the composition is a monomorphism which is locally of finite type. On the other hand, the composition is integral as the map on underlying reduced affine schemes is a closed immersion by the above and hence universally closed (use Morphisms, Lemma 29.44.7). Thus the ring map
\mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times _ X V)
is an epimorphism which is integral of finite type hence finite hence surjective (use Morphisms, Lemma 29.44.4 and Algebra, Lemma 10.107.6).
\square
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