= Solution
A <finite morphism> is a <finite type morphism>. It is also a <separated morphism>: on an <affine scheme> chart $\operatorname{Spec}B\to\operatorname{Spec}A$, its <diagonal morphism> is represented by the surjective multiplication <ring homomorphism> $B\otimes_AB\to B$, and is therefore a <closed immersion> by Question 1.
We prove the lifting condition in the <valuative criterion for properness>. Choose an <affine open subset> of $Y$ containing the image of the closed point of $\operatorname{Spec}V$. Every point of $\operatorname{Spec}V$ is a generalization of its closed point, so the whole map factors through this <affine open subset>. Its inverse image under the <finite morphism> is $\operatorname{Spec}B$, with $B$ a finitely generated $A$-<module>. Every $b\in B$ is integral over $A$: apply the <determinant trick> to multiplication by $b$ on a finite set of <module> generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.
In the given valuative square, we have compatible <ring homomorphisms> $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a <valuation ring> gives $x^{-1}\in V$, and it lies in the <maximal ideal> since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that <maximal ideal>, a contradiction. This proves that <valuation rings are integrally closed> and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving
$$
\boxed{\text{finite morphisms are proper}.}
$$
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