With or without using Chevalley's theorem, show that if $X\to \text{Spec }k$ is a quasifinite morphism (according to Vakil, this means finite type morphism + finite fibers), then the morphism is actually finite.
This question was asked before (in FOAG exercise 7.4.D.), but I don't quite understand the solution:
First, even if solve the case where $X$ is affine, how do we show that the claim holds in general for any scheme $X$?
Moreover, the accepted solution suggests that $X=\text{Spec }A$ is integral ("consider the generic point of $X$"), that the morphism $\text{Spec }A \to\text{Spec }k[x]$ induced by the inclusion $k[x]\subset A$ is dominant, and that the condition on finite fibers implies that the generic point of $X$ is constructible, but none of these are particularly clear to me.
Do we need some sort of requirement that the fiber is discrete (this would make, for instance, the last claim easy to verify)?