Proof.
A finite morphism is proper according to Morphisms of Spaces, Lemma 67.45.9. A finite morphism is quasi-finite according to Morphisms of Spaces, Lemma 67.45.8. A quasi-finite morphism has discrete fibres X_ k, see Morphisms of Spaces, Lemma 67.27.5. Hence a finite morphism is proper and has discrete fibres X_ k.
Assume f is proper with discrete fibres X_ k. We want to show f is finite. In fact it suffices to prove f is affine. Namely, if f is affine, then it follows that f is integral by Morphisms of Spaces, Lemma 67.45.7 whereupon it follows from Morphisms of Spaces, Lemma 67.45.6 that f is finite.
To show that f is affine we may assume that Y is affine, and our goal is to show that X is affine too. Since f is proper we see that X is separated and quasi-compact. We will show that for any coherent \mathcal{O}_ X-module \mathcal{F} we have H^1(X, \mathcal{F}) = 0. This implies that H^1(X, \mathcal{F}) = 0 for every quasi-coherent \mathcal{O}_ X-module \mathcal{F} by Lemmas 69.15.1 and 69.5.1. Then it follows that X is affine from Proposition 69.16.7. By Lemma 69.22.8 we conclude that the stalks of R^1f_*\mathcal{F} are zero for all geometric points of Y. In other words, R^1f_*\mathcal{F} = 0. Hence we see from the Leray Spectral Sequence for f that H^1(X , \mathcal{F}) = H^1(Y, f_*\mathcal{F}). Since Y is affine, and f_*\mathcal{F} is quasi-coherent (Morphisms of Spaces, Lemma 67.11.2) we conclude H^1(Y, f_*\mathcal{F}) = 0 from Cohomology of Schemes, Lemma 30.2.2. Hence H^1(X, \mathcal{F}) = 0 as desired.
\square
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