Let $(R,\mathfrak{m})$ be a local ring, denote by $R \rightarrow R^h$ its henselization. Write $S = \text{Spec} R$ and $S^h = \text{Spec} R^h$. Is it true that the diagonal morphism $\Delta \colon S^h \rightarrow S^h \times_S S^h$ is an open immersion? I am reading the chapter by Orgogozo and Vidal in ["Courbes semi-stables et groupe fondamental en geometrie algebrique"](https://link.springer.com/book/9783764363086), where this is used (without proof) in the proof of Lemme 5.4. 

I have tried the following. Write 
$$R^h = \varinjlim_{(A,\mathfrak{q})} A,$$
with the limit running over the pairs $(A,\mathfrak{q})$, where $A$ is an étale algebra over $R$, and $\mathfrak{q} \in \text{Spec} A$ is such that $\kappa(\mathfrak{q}) = R/\mathfrak{m}$ ([TAG 04GN](https://stacks.math.columbia.edu/tag/04GN)). Then 
$$ S = \varprojlim_{(A,\mathfrak{q})} \text{Spec} A,$$
and
$$S^h \times_S S^h = (\varprojlim_{(A,\mathfrak{q})} \text{Spec} A) \times_R (\varprojlim_{(A,\mathfrak{q})} \text{Spec} A) = \varprojlim_{(A,\mathfrak{q})} (\text{Spec} A \times_R \text{Spec} A).$$
Now the diagonal morphism $\Delta \colon S^h \rightarrow S^h \times_S S^h$ is the $\varprojlim$ of the individual diagonal morphisms $\text{Spec} A \rightarrow \text{Spec} A \times_R \text{Spec} A$. Each of them is an open immersion, because $A$ is étale over $R$, in particular, $A$ is unramified over $R$, and hence we can apply [TAG 02GE](https://stacks.math.columbia.edu/tag/02GE). Can I somehow deduce from this that $\Delta$ is itself an open immersion? I tried to apply the theory in EGA IV.3, Section 8, but it is not obvious to me how the setup there is applicable to this problem. 
Any help would be greatly appreciated!