= Solution
The composite $g\circ f$ is <proper morphism>[proper], hence separated. If two lifts $\operatorname{Spec}R\to X$ solve a valuation-ring lifting problem for $f$, they also solve the corresponding problem for $g\circ f$. The <valuative criterion for separatedness> for $g\circ f$ makes them equal, so $f$ is separated.
It remains to prove existence. Start with a square
$$
\begin{array}{ccc}
\operatorname{Spec}K&\xrightarrow{u}&X\\
\downarrow&&\downarrow f\\
\operatorname{Spec}R&\xrightarrow{v}&Y.
\end{array}
$$
After composing the lower map with $g$, properness of $g\circ f$ gives a lift $a:\operatorname{Spec}R\to X$ over $Z$ whose generic restriction is $u$. The two maps $f\circ a$ and $v$ from $\operatorname{Spec}R$ to $Y$ agree on $\operatorname{Spec}K$ and have the same composite with $g$. Since $g$ is separated, its valuative uniqueness criterion gives $f\circ a=v$. Hence $a$ is the required lift for $f$.
The morphism $f$ is of finite type by hypothesis, and it is separated and satisfies valuative existence. The <valuative criterion for properness> therefore proves that $f$ is proper.
Solved by gpt-5.6-sol high.
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