Let $X$ be an integral Noetherian scheme such that $\mathcal{O}_X(X)$ is Noetherian (the assumption is not redundant https://math.stackexchange.com/a/818591/691994).
Is the affinization map $X\rightarrow \mathrm{Spec}\:\mathcal{O}_X(X)$ open?
I can not even rule out the possibility that the image of $X$ is a proper closed subset.
Partial results: the statement is true
- for quasi-affine schemes (https://stacks.math.columbia.edu/tag/01P9);
- for schemes $X$ such that there is a proper flat morphism with geometrically integral fibers $X\rightarrow \mathrm{Spec}\:R$ where $R$ is Noetherian. This is because the canonical map $R\rightarrow \mathcal{O}_X(X)$ is an isomorphism (https://math.stackexchange.com/a/475606/691994) so the affinization map is the structure morphism and flat morphism of finite type to a Noetherian scheme is open (https://stacks.math.columbia.edu/tag/01TX, https://stacks.math.columbia.edu/tag/01UA).