Let $R$ be a valuation ring with maximal ideal $M$ and $L$ its residue field. We let $k$ be the fractions field which we assume to be algebraically closed. Let $\mathcal{L}$ be a closed subset of $\mathbb{P}^n_R$ defined by $r+1$ linear forms with coefficients in $R$. Let $\tau: \mathbb{P}^n_R \backslash \mathcal{L} \to \mathbb{P}^r_R$ be the projection.
Let $p: \mathbb{P}^n_R \to \operatorname{Spec}R$, and $i_b: \operatorname{Spec}L \to \operatorname{Spec}R $ where the image is $b$, the closed point of $\operatorname{Spec}R$. It can be shown that the fiber of $p$ over $b$ is $\mathbb{P}^n_L$ and let $j: \mathbb{P}^n_L \to \mathbb{P}^n_R$ be the morphism we obtain from the universal property of the fiber products. It turns out that $j$ is a closed immersion and through this morphism we can view $\mathbb{P}_L^n$ inside $\mathbb{P}_R^n$. Similarly, by considering the fiber product of $p$ over the generic point, we obtain $i: \mathbb{P}^n_k \to \mathbb{P}^n_R$.
Let $Z$ be a closed subset of $\mathbb{P}^n_k$ and $\mathcal{Z}$ the closure of $i(Z)$ inside $\mathbb{P}^n_R$. Assume $\mathcal{Z}$ is disjoint from $\mathcal{L}$. Then it is proved that $\tau: Z \to \mathbb{P}_R^r$ is a finite surjective morphism.
I understand the proof until the end of page 181 (in the old version), where he deduces that $\tilde{x} \in Z$. He then claims that $\tilde{x}$ is a closed point of $Z$, because $\tau_k(\tilde{x})$, where $\tau_k: Z \to \mathbb{P}^r_k$, is a closed point. I am not seeing how this is deduced knowing that $\tau_k(\tilde{x})$ is a closed point. I would appreciate if some could clarify this. Thank you!
$\tau_k^{-1}(\tau_k(\widetilde{x}))$ is closed inside $Z$, as it is the preimage of a closed subset. On the other hand, the fiber of a finite morphism over any point is finite discrete - thus every point of the fiber is closed in the fiber. As the underlying topological space of the scheme-theoretic fiber is the same as the topological fiber (ref), we have that $\{\widetilde{x}\}$ is a closed subset of $\tau_k^{-1}(\tau_k(\widetilde{x}))$ which is again a closed subset of $Z$. As a closed subset of a closed subset is again closed, we have the result.