The Stacks project

Proof. The scheme $U$ is quasi-affine because integral morphisms are affine, affine morphisms are quasi-affine, a scheme is quasi-affine if and only if the structure morphism to $\mathop{\mathrm{Spec}}(\mathbf{Z})$ is quasi-affine, and compositions of quasi-affine morphisms are quasi-affine. The first two statements follow immediately from the definition and the third is Morphisms, Lemma 29.13.4. Set $U' = X \times _{\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))} \mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ and consider the extended diagram

The morphism $j$ is closed by Morphisms, Lemma 29.41.7 combined with the fact that an integral morphism is universally closed (Morphisms, Lemma 29.44.7) and the fact that the vertical arrows are in the diagram are separated. On the other hand, $j$ is open because the horizontal arrows in the diagram of the lemma are open by Properties, Lemma 28.18.4. Thus $j$ identifies $U$ with an open and closed subscheme of $U'$. If $U \not= U'$ then $U$ isn't dense in $U'$ and a fortiori not dense in the spectrum of $\Gamma (U, \mathcal{O}_ U)$. However, the scheme theoretic image of $U$ in $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ is $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ because any ideal in $\Gamma (U, \mathcal{O}_ U)$ cutting out a closed subscheme through which $U$ factors would have to be zero. Hence $U$ is dense in $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{O}_ U))$ for example by Morphisms, Lemma 29.6.3. Thus $U = U'$ and we win. $\square$