This is related to Ueno's Algebraic Geometry 2, Thm 5.9 Valuative Criterion proof.
Let $f:X\to Y$ be a finite type morphism between $X$ noetherian scheme and $Y$ scheme. Assume $f$ is proper and hence $f$ is separated.
I want to see for any valuation ring $R$ with $Frac(R)=K$ s.t. given commutative diagram $Spec(R)\to Y$ and $Spec(K)\to X$ inducing unique $Spec(R)\to X$ lifting s.t. overall triangles of diagram commute.
Suppose $g,g'$ are 2 such liftings. Let $\eta_0,\eta'_0$ be closed points of $Spec(R)$ and $\zeta_0,\zeta'_0$ be corresponding images under lift $g,g'$ to $X$. It is clear that diagonal map induced by $f:X\to Y$ for $X\to X\times_YX$ is closed by $f$ separated. Note that $Spec(K)\to Spec(R)$ is identifying the generic point and $g,g'$ agrees over image of $Spec(K)$ map to $X$.
Now $Spec(R)\to X\times_YX$ will have generic point image lying on diagonal. It suffices to check closed point $Spec(R)$ is on diagonal as well. Note that image of $X$ under diagonal map is closed. Since $X$ is noetherian scheme, $X\to X\times_YX$ has closed image iff $Im(X\to X\times_YX)$ is closed under specialization. If that is the case, then $g=g'$. Denote $\eta_1$ as image of $Spec(K)\to X$.
$\textbf{Q:}$ Why is image of $(\eta_0,\eta'_0)\in\overline{(\eta_1,\eta_1)}\in X\times_YX$? In other words, why $(\eta_0,\eta'_0)$ is specialization of $(\eta_1,\eta_1)$? It seems that the book is using continuity of the map $Spec(R)\to X\times_YX$ to conclude $Im\subset\overline{(\eta_1,\eta_1)}$. In other words, consider continuous map $h:X\to Y$ and say $x\in X$ I have $h(\overline{x})\subset \overline{h(x)}$
Best Answer
First note that any valuation ring $R$ is local, with maximal ideal $\mathfrak{m} = \{r \in R | v(r) > 0\}$. So $\operatorname{Spec}(R)$ has only one closed point, i.e. $\eta_0 = \eta_0'$.
Then $f^{-1}(\overline{(\eta_1, \eta_1)}) \subset \operatorname{Spec }(R)$ is a closed set. Any closed set in an affine scheme contains closed points, but there is only one closed point in $\operatorname{Spec }(R)$, namely $\eta_0$. But $\eta_0 \in f^{-1}(\overline{(\eta_1, \eta_1)})$ if and only if $(\eta_0, \eta_0) = f(\eta_0) \in \overline{(\eta_1, \eta_1)}$.