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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