A morphism of triples $(S, \mathfrak q, \phi) \to (S', \mathfrak q', \phi')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi^{-1}(\mathfrak q') = \mathfrak q$ and such that $\phi' \circ \varphi = \phi$. Let us show that the category of pairs is filtered, see Categories, Definition \ref{categories-definition-directed}. The category contains the triple $(R, \mathfrak p, \kappa(\mathfrak p) \subset \kappa^{sep})$ and hence is not empty, which proves part (1) of Categories, Definition \ref{categories-definition-directed}. Suppose that $(S, \mathfrak q, \phi)$ and $(S', \mathfrak q', \phi')$ are two triples. Note that $\mathfrak q$, resp.\ $\mathfrak q'$ correspond to primes of the fibre rings $S \otimes \kappa(\mathfrak p)$, resp.\ $S' \otimes \kappa(\mathfrak p)$ with residue fields finite separable over $\kappa(\mathfrak p)$ and $\phi$, resp.\ $\phi'$ correspond to maps into $\kappa^{sep}$. Hence this data corresponds to $\kappa(\mathfrak p)$-algebra maps $$ \phi : S \otimes_R \kappa(\mathfrak p) \longrightarrow \kappa^{sep}, \quad \phi' : S' \otimes_R \kappa(\mathfrak p) \longrightarrow \kappa^{sep}. $$ Set $S'' = S \otimes_R S'$. Combining the maps the above we get a unique $\kappa(\mathfrak p)$-algebra map $$ \phi'' = \phi \otimes \phi' : S'' \otimes_R \kappa(\mathfrak p) \longrightarrow \kappa^{sep} $$ whose kernel corresponds to a prime $\mathfrak q'' \subset S''$ lying over $\mathfrak q$ and over $\mathfrak q'$, and whose residue field maps via $\phi''$ to the compositum of $\phi(\kappa(\mathfrak q))$ and $\phi'(\kappa(\mathfrak q'))$ in $\kappa^{sep}$. The ring map $R \to S''$ is \'etale by Lemma \ref{lemma-etale}. Hence $(S'', \mathfrak q'', \phi'')$ is a triple dominating both $(S, \mathfrak q, \phi)$ and $(S', \mathfrak q', \phi')$. This proves part (2) of Categories, Definition \ref{categories-definition-directed}. Next, suppose that $\varphi, \psi : (S, \mathfrak q, \phi) \to (S', \mathfrak q', \phi')$ are two morphisms of pairs. Then $\varphi$, $\psi$, and $S' \otimes_R S' \to S'$ are \'etale ring maps by Lemma \ref{lemma-map-between-etale}. Consider $$ S'' = (S' \otimes_{\varphi, S, \psi} S') \otimes_{S' \otimes_R S'} S' $$ Arguing as above (base change of \'etale maps is \'etale, composition of \'etale maps is \'etale) we see that $S''$ is \'etale over $R$. The fibre ring of $S''$ over $\mathfrak p$ is $$ F'' = (F' \otimes_{\varphi, F, \psi} F') \otimes_{F' \otimes_{\kappa(\mathfrak p)} F'} F' $$ where $F', F$ are the fibre rings of $S'$ and $S$. Since $\varphi$ and $\psi$ are morphisms of triples the map $\phi' : F' \to \kappa^{sep}$ extends to a map $\phi'' : F'' \to \kappa^{sep}$ which in turn corresponds to a prime ideal $\mathfrak q'' \subset S''$. The canonical map $S' \to S''$ (using the right most factor for example) is a morphism of triples $(S', \mathfrak q', \phi') \to (S'', \mathfrak q'', \phi'')$ which equalizes $\varphi$ and $\psi$. This proves part (3) of Categories, Definition \ref{categories-definition-directed}. Hence we conclude that the category is filtered. \medskip\noindent We still have to show that the colimit $R_{colim}$ of the system is equal to the strict henselization of $R_{\mathfrak p}$ with respect to $\kappa^{sep}$. To see this note that the system of triples $(S, \mathfrak q, \phi)$ contains as a subsystem the pairs $(S, \mathfrak q)$ of Lemma \ref{lemma-henselization-different}. Hence $R_{colim}$ contains $R_{\mathfrak p}^h$ by the result of that lemma. Moreover, it is clear that $R_{\mathfrak p}^h \subset R_{colim}$ is a directed colimit of \'etale ring extensions. It follows that $R_{colim}$ is henselian by Lemmas \ref{lemma-finite-over-henselian} and \ref{lemma-colimit-henselian}. Finally, by Lemma \ref{lemma-make-etale-map-prescribed-residue-field} we see that the residue field of $R_{colim}$ is equal to $\kappa^{sep}$. Hence we conclude that $R_{colim}$ is strictly henselian and hence equals the strict henselization of $R_{\mathfrak p}$ as desired. Some details omitted.