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I want to show:

Let $R$ be a Dedekind domain with fraction field $K$. Let $X$ be a scheme and $X \to \operatorname{Spec}R$ a proper morphism. Show that the natural map $$X(\operatorname{Spec}R)\to X (\operatorname{Spec}K)$$ is a bijection. Here $X(Y):=\text{Hom}(Y,X)$.

I know that if R is a valuation ring, then this is exactly the valuative criterion for properness. So I consider the localization of $R$ at every maximal ideal $m$, try to show that the data of $X(\operatorname{Spec}R)$ is equivalent to the data of $X(\operatorname{Spec}R_m)$ for all $m$ with some conditions. But it seems not so clear.

Any hint or reference would be helpful.

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    Do you see that the only part which needs a proof is surjectivity? If so, given a morphism $\mathrm{Spec}\, K\to X$ (over $R$), take the image and its closure in $X$. Show that the map from this closure to $\mathrm{Spec}\, R $ is an isomorphism. – 2017-01-23
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    @Mohan Yeah, I know how to show injectivity. Could you explain why the map from its closure is an isomorphism? – 2017-01-23
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    Let $Z$ be the closure, then it is irreducible, being the closure of a singleton. The map from $Z$ to $R$ is birational and thus its dimension is one. So, the fibers are finite and since the map is proper, the map is in fact finite. Thus, $Z$ is affine, and if $Z=\mathrm{Spec}\, S$, then $R\to S$ is finite and birational. Since $R$ is a Dedekind domain, this shows $R=S$. – 2017-01-23

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