= Solution
Let $S'\to S$ be a <morphism of schemes> and set $X'=X\times_SS'$, with projection $f':X'\to S'$. Given compatible <morphisms of schemes> $u':\operatorname{Spec}K\to X'$ and $v':\operatorname{Spec}D\to S'$, compose them with the projections to obtain a valuative problem for $f$. An extension $\ell:\operatorname{Spec}D\to X$ for that problem and the prescribed $v'$ determine a unique <morphism of schemes>
$$
\ell'=(\ell,v'):\operatorname{Spec}D\longrightarrow X\times_SS'.
$$
The <universal property of a fibre product> shows that $\ell'$ extends $u'$ and lies over $v'$. Conversely, every extension $\ell'$ has such a first component $\ell$. Thus the sets of extensions in the two problems are in <bijection>.
If $f$ is a <universally closed morphism>, it is a <quasi-compact morphism> and satisfies existence. The allowed stability of <quasi-compact morphisms> under <base change> and (a)(i) prove that $f'$ is a <universally closed morphism>. If $f$ is a <separated morphism>, it is a <quasi-separated morphism> and satisfies uniqueness. Stability of <quasi-separated morphisms> under <base change> and (a)(ii) prove that $f'$ is a <separated morphism>.
If $f$ is a <proper morphism>, the same <bijection> supplies existence and uniqueness. The permitted stability under <base change> of <quasi-compact morphisms>, <quasi-separated morphisms> and <locally of finite type> supplies all remaining hypotheses of (a)(iii). Hence
$$
\boxed{\begin{aligned}
f\text{ universally closed}&\Longrightarrow f'\text{ universally closed},\\
f\text{ separated}&\Longrightarrow f'\text{ separated},\\
f\text{ proper}&\Longrightarrow f'\text{ proper}.
\end{aligned}}
$$
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