Strict henselian etale sections

Let $x\in H(k)$ and choose a standard etale chart $R[T]\mathrm{/}(P)$ localized at $g$ around $x$ as in [F1]. The image of the chart in $\mathrm{Spec}R$ is an open neighbourhood of the image point, which is the closed point of the local scheme $\mathrm{Spec}R$; hence it is all of $\mathrm{Spec}R$. Write $a$ for the residue class of $T$ at $x$; it is a simple root of the monic polynomial $P$ because $P^{\mathrm{\prime }}$ is invertible on the chart. By the simple-root lifting criterion [F2] there is a lift $a\in R$ with $P(a)=0$ and $g(a)\mathrm{\notin }\mathfrak{m}$, so evaluation at $a$ defines an $R$-section of the chart and hence of $H$ reducing to $x$. Thus $H(R)\to H(k)$ is surjective.

Two sections $s,t$ of $H$ with equal reduction have equalizer $\mathrm{Eq}(s,t)\subseteq \mathrm{Spec}R$; the diagonal of an etale morphism is an open immersion, so it is open, and $H$ separated over $R$ makes it closed, while it contains the closed point by hypothesis; since $\mathrm{Spec}R$ is connected (it is a local scheme), the equalizer is all of $\mathrm{Spec}R$, so $s=t$. Hence reduction $H(R)\to H(k)$ is injective, and with step 1.1 it is bijective; this proves (a).

For a section $s:\mathrm{Spec}R\to H$, its image is open, being the base change of the etale diagonal, and closed because $s$ is a closed immersion as a section of the separated morphism $H\to \mathrm{Spec}R$. Two distinct sections have disjoint images: their equalizer is open and closed by the same argument as in step 2.1, and it is empty because it is a proper closed subset of the connected scheme $\mathrm{Spec}R$ (it misses the closed point since the sections have distinct reductions by the bijection of step 2.1). Hence the images of the $d$ sections, one for each point of the finite set $H(k)$, form $d$ disjoint open and closed subschemes each isomorphic to $\mathrm{Spec}R$ via $s$.

Since $H$ is etale and finite type over the field $k$, the closed fibre $H_k$ is a disjoint union of finitely many copies of $\mathrm{Spec}k$ (finite separable extensions of the separably closed field $k$ are trivial), so the closed fibre is covered by the closed points $H(k)$, and $H_k\subseteq H^{\mathrm{f}\mathrm{i}\mathrm{n}}$. Hence $H^{\mathrm{f}\mathrm{i}\mathrm{n}}$, the disjoint union of the $d$ section images, is finite etale over $R$, and every section of $H$ meets $H_k$ and therefore lies in $H^{\mathrm{f}\mathrm{i}\mathrm{n}}$; the complement $H\setminus H^{\mathrm{f}\mathrm{i}\mathrm{n}}$ has empty closed fibre and admits no $R$-section. This proves (b).