Let $f: C \to D$ be a proper, non constant scheme morphism of curves. (a curve is for me a $1$-dimensional, separated $k$-scheme of finite type).
Assume futhermore that $C$ and $D$ are irreducible and proper.
Let $b \in D$ an arbitrary closed point of $D$. My question is how to see that the fiber $f^{-1}(b)$ is a finite set?
My considerations:
Since the property "finite type" is stable under base change we deduce that the scheme structure $C \times_D \kappa(b)$ of the fiber $f^{-1}(b)$ is a $\kappa(b)$-scheme of finite type. Therefore the ring of grobal sections of $C \times_D \kappa(b)$ is Noetherian by Hilbert's Basissatz.
Futhermore $f^{-1}(b)$ is discrete and a union of closed points. But here I don't see why the beeing Noetherian property for the ring $O_{C \times_D \kappa(b)}(C \times_D \kappa(b))$ imply the Noether ascending property for $C \times_D \kappa(b)$ as topological space. The problem is that $C \times_D \kappa(b)$ isn't affine.
Another approach would be to show that every complement of an non empty open set in $C$ is finite but I'm not sure why it should here hold. I can only say that every open set of $C$ is dense but not more.
If there were an infinite fiber, then it would have an infinite irreducible component. That is, we have an infinite closed irreducible set inside of the curve $C$. This is impossible for dimension reasons unless that component is the whole curve. But the morphism is not constant, so the component can’t be the whole curve.