Consider the category of pairs $(S, \mathfrak q)$ where $R \to S$ is an \'etale ring map, and $\mathfrak q$ is a prime of $S$ lying over $\mathfrak m$ with $\kappa = \kappa(\mathfrak q)$. A morphism of pairs $(S, \mathfrak q) \to (S', \mathfrak q')$ is given by an $R$-algebra map $\varphi : S \to S'$ such that $\varphi^{-1}(\mathfrak q') = \mathfrak q$. We set $$ R^h = \colim_{(S, \mathfrak q)} S. $$ Let us show that the category of pairs is filtered, see Categories, Definition \ref{categories-definition-directed}. The category contains the pair $(R, \mathfrak m)$ and hence is not empty, which proves part (1) of Categories, Definition \ref{categories-definition-directed}. For any pair $(S, \mathfrak q)$ the prime ideal $\mathfrak q$ is maximal with residue field $\kappa$ since the composition $\kappa \to S/\mathfrak q \to \kappa(\mathfrak q)$ is an isomorphism. Suppose that $(S, \mathfrak q)$ and $(S', \mathfrak q')$ are two objects. Set $S'' = S \otimes_R S'$ and $\mathfrak q'' = \mathfrak qS'' + \mathfrak q'S''$. Then $S''/\mathfrak q'' = S/\mathfrak q \otimes_R S'/\mathfrak q' = \kappa$ by what we said above. Moreover, $R \to S''$ is \'etale by Lemma \ref{lemma-etale}. This proves part (2) of Categories, Definition \ref{categories-definition-directed}. Next, suppose that $\varphi, \psi : (S, \mathfrak q) \to (S', \mathfrak q')$ 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 $$ S'' = (S' \otimes_{\varphi, S, \psi} S') \otimes_{S' \otimes_R S'} S' $$ with prime ideal $$ \mathfrak q'' = (\mathfrak q' \otimes S' + S' \otimes \mathfrak q') \otimes S' + (S' \otimes_{\varphi, S, \psi} S') \otimes \mathfrak q' $$ Arguing as above (base change of \'etale maps is \'etale, composition of \'etale maps is \'etale) we see that $S''$ is \'etale over $R$. Moreover, the canonical map $S' \to S''$ (using the right most factor for example) equalizes $\varphi$ and $\psi$. This proves part (3) of Categories, Definition \ref{categories-definition-directed}. Hence we conclude that $R^h$ consists of triples $(S, \mathfrak q, f)$ with $f \in S$, and two such triples $(S, \mathfrak q, f)$, $(S', \mathfrak q', f')$ define the same element of $R^h$ if and only if there exists a pair $(S'', \mathfrak q'')$ and morphisms of pairs $\varphi : (S, \mathfrak q) \to (S'', \mathfrak q'')$ and $\varphi' : (S', \mathfrak q') \to (S'', \mathfrak q'')$ such that $\varphi(f) = \varphi'(f')$. \medskip\noindent Suppose that $x \in R^h$. Represent $x$ by a triple $(S, \mathfrak q, f)$. Let $\mathfrak q_1, \ldots, \mathfrak q_r$ be the other primes of $S$ lying over $\mathfrak m$. Then $\mathfrak q \not \subset \mathfrak q_i$ as we have seen above that $\mathfrak q$ is maximal. Thus, since $\mathfrak q$ is a prime ideal, we can find a $g \in S$, $g \not \in \mathfrak q$ and $g \in \mathfrak q_i$ for $i = 1, \ldots, r$. Consider the morphism of pairs $(S, \mathfrak q) \to (S_g, \mathfrak qS_g)$. In this way we see that we may always assume that $x$ is given by a triple $(S, \mathfrak q, f)$ where $\mathfrak q$ is the only prime of $S$ lying over $\mathfrak m$, i.e., $\sqrt{\mathfrak mS} = \mathfrak q$. But since $R \to S$ is \'etale, we have $\mathfrak mS_{\mathfrak q} = \mathfrak qS_{\mathfrak q}$, see Lemma \ref{lemma-etale-at-prime}. Hence we actually get that $\mathfrak mS = \mathfrak q$. \medskip\noindent Suppose that $x \not \in \mathfrak mR^h$. Represent $x$ by a triple $(S, \mathfrak q, f)$ with $\mathfrak mS = \mathfrak q$. Then $f \not \in \mathfrak mS$, i.e., $f \not \in \mathfrak q$. Hence $(S, \mathfrak q) \to (S_f, \mathfrak qS_f)$ is a morphism of pairs such that the image of $f$ becomes invertible. Hence $x$ is invertible with inverse represented by the triple $(S_f, \mathfrak qS_f, 1/f)$. We conclude that $R^h$ is a local ring with maximal ideal $\mathfrak mR^h$. The residue field is $\kappa$ since we can define $R^h/\mathfrak mR^h \to \kappa$ by mapping a triple $(S, \mathfrak q, f)$ to the residue class of $f$ modulo $\mathfrak q$. \medskip\noindent We still have to show that $R^h$ is henselian. Namely, suppose that $P \in R^h[T]$ is a monic polynomial and $a_0 \in \kappa$ is a simple root of the reduction $\overline{P} \in \kappa[T]$. Then we can find a pair $(S, \mathfrak q)$ such that $P$ is the image of a monic polynomial $Q \in S[T]$. Since $S \to R^h$ induces an isomorphism of residue fields we see that $S' = S[T]/(Q)$ has a prime ideal $\mathfrak q' = (\mathfrak q, T - a_0)$ at which $S \to S'$ is standard \'etale. Moreover, $\kappa = \kappa(\mathfrak q')$. Pick $g \in S'$, $g \not \in \mathfrak q'$ such that $S'' = S'_g$ is \'etale over $S$. Then $(S, \mathfrak q) \to (S'', \mathfrak q'S'')$ is a morphism of pairs. Now that triple $(S'', \mathfrak q'S'', \text{class of }T)$ determines an element $a \in R^h$ with the properties $P(a) = 0$, and $\overline{a} = a_0$ as desired.