Preimage of a constructible set in spectrum of a subring

Question

While working through a proof of this paper, at the beginning of page 42, the author seems to claim the following is true:

Let $R\subset S$ be rings, where $R$ is a finite type algebra over $\mathbb F_p$. Consider the associated map of the prime spectra $$ \varphi:\text{Spec}(S)\rightarrow \text{Spec}(R). $$ Suppose that $K\subset \text{Spec}(R)$ is a constructible subset such that $\varphi^{-1}(K)=\varnothing$. Prove that there exists an $R\subset R^{'}\subset S$, such that $R'$ is a finite type $R-$algebra and that if $$ \psi:\text{Spec}(R')\rightarrow \text{Spec}(R). $$ is the associated map of spectra, then $\psi^{-1}(K)=\varnothing$.

I believe that I have an argument for the case when $K$ is a finite subset. One could think of $S$ as a direct limit of its finitely generated $R$-subalgebras and therefore $Spec(S)$ should equal an inverse limit of the spectra of the finitely generated $R$-subalgebras. For each prime in $K$, choose a finitely generated $R$-subalgebra where it does not have a preimage, and the rest is clear. However, I don't know what to do for the general case.

Answer 1

See for example Tag 05F4. If you click through you find the following argument: write $K$ as the image of a map $\text{Spec}(A) \to \text{Spec}(R)$ for some ring map $R \to A$; this is where you use that $K$ is constructible. Then $\varphi^{-1}(K) = \emptyset$ means that $A \otimes_R S$ is the zero ring, in other words, $1 = 0$ in $A \otimes_R S$. Since $S = \text{colim} R'$ we find that $1 = 0$ already in some $A \otimes_R R'$.