Open immersions are etale

The local model is standard 'etale. Let $A$ be a ring and $g\in A$; the $A$-algebra map $A[T]\to A$ with $T\mapsto g$ is surjective with kernel $(T-g)$ by the first isomorphism theorem for rings, so $A[T]\mathrm{/}(T-g)\cong A$, and localising this isomorphism at the image $\bar{T}=g$ of $T$ gives $(A[T]\mathrm{/}(T-g){)}_T\cong A_g$. The polynomial $T-g$ is monic and its formal derivative is the constant $1$, whose image in $A_g$ is a unit; hence $A_g$ is standard 'etale over $A$ by [F2], and since it is finitely presented over $A$ the map $\mathrm{Spec}{A}_g\to \mathrm{Spec}A$ is 'etale, and by [F1] it is the open immersion onto $D(g)$.

An open immersion is locally the localisation model. Let $j\text{\hspace{0.17em}⁣}:U\to X$ be an open immersion and $u\in U$. By [F3] $j$ identifies $U$ with the open subscheme $j(U)\subseteq X$. Choose an affine chart $\mathrm{Spec}A\subseteq X$ containing $j(u)$; then $V:=j(U)\cap \mathrm{Spec}A$ is an open neighbourhood of $j(u)$ in $\mathrm{Spec}A$, so by [F1] there is $g\in A$ with $j(u)\in D(g)\subseteq V$. Again by [F3] the map $j^{-1}(D(g))\to D(g)$ is an isomorphism, and $D(g)=\mathrm{Spec}{A}_g$ by [F1]; on these open neighbourhoods the morphism $j$ is therefore the localisation map $\mathrm{Spec}{A}_g\to \mathrm{Spec}A$ of step 1.1, up to the identifications, and it is 'etale there. By the germ property [F4] $j$ is 'etale at $u$; the same applies to the empty immersion vacuously, since then there is no point $u$.

Conclusion. Since every point of $U$ is covered by step 2.1, the open immersion $j$ is 'etale. Steps 1.1 and 2.1 use only the standard 'etale presentation, the localisation identification of spectra, the first isomorphism theorem and the local nature of 'etaleness, so the argument is choice-free and no Axiom of Choice is assumed or used.