Let $S$ be a scheme and $S_0 \subset S$ a closed subscheme. Then $(S, S_0)$ is said to be a Henselian couple if for every finite $X \rightarrow S$, setting $X_0 := X\times_S S_0$, the natural map from the clopen subsets of $X$ to those of $X_0$ is bijective. An example is a Henselian local (or semilocal) ring and its maximal (or radical) ideal; in fact, this is one of the equivalent definitions of a Henselian local ring. For this choice, EGA IV 18.5.15 proves that the base change functor is an equivalence of categories between finite etale $S$schemes and finite etale $S_0$schemes. More generally, base changing this choice along a proper $X \rightarrow S$ one obtains other examples of Henselian couples for which the same conclusion holds; these are of relevance for the proper base change theorem in SGA 4, Exp. XII. My question is: does the conclusion about the equivalence of categories of finite etale schemes hold for arbitrary Henselian couples or are there counterexamples? EGA IV 18.5.16 remarks that this is not known, but what is the status of the question today?
At least when $S$ is affine, this follows from work of Elkik. However, in this case, it seems that one can also argue directly as explained below.
[ It's surprising this argument is not in EGA, so please read critically.]
Say $S = \mathrm{Spec}(A)$ and $S_0 = \mathrm{Spec}(A/I)$ with $A$ henselian along $I$. I will use a subscript $0$ to denote passage to the fibre over $S_0$ on $S$schemes, and implicitly use that any finite $S$schemes $T$ is henselian along $T_0$.
The goal is to show that every finite etale map $Y_0 \to S_0$ lifts to a finite etale map $Y \to X$. (The full faithfulness part is easy from basic properties about henselian pairs.)
By limit arguments, one may assume that $A$ is noetherian for convenience. (Indeed, we may write any pair $(B,J)$ can be written as a filtered colimit of pairs $(B_i,J_i)$ with $B_i$ noetherian. Now observe that the inclusion of henselian pairs in all pairs is fully faithful, and has a left adjoint given by henselization.)
Begin by picking an etale map $Y' \to S$ extending $Y_0 \to S_0$ (this is elementary, see http://stacks.math.columbia.edu/tag/04D1). Using Zariski's main theorem for $Y' \to S$, we can find a factorization
$$Y' \stackrel{j}{\to} \overline{Y'} \stackrel{\pi}{\to} S$$
with $j$ an open immersion, $\pi$ a finite morphism, $Y' \to S$ etale and $Y'_0 = Y_0$. Restricting $j$ to $S_0$ gives an open immersion $j_0:Y_0 \to \overline{Y'}_0$ of finite $S_0$schemes, so $j_0(Y_0)$ is clopen in $\overline{Y'}_0$. As $\overline{Y'}$ is henselian along $\overline{Y'}_0$, there is a unique clopen subset $\overline{Z} \subset \overline{Y'}_0$ such that $\overline{Z}_0 = Y_0$ as clopen subsets of $\overline{Y'}_0$. Intersecting the above factorization with $\overline{Z}$ gives a factorisation
$$ Z \stackrel{k}{\to} \overline{Z} \stackrel{f}{\to} S$$
such that $k$ is an open immersion, $f$ is finite, $Z := \overline{Z} \cap Y'$ is etale over $S$, and $Y_0 = Z_0 = \overline{Z}_0$ (as $Y_0 = \overline{Z}_0$ and $Y_0 \subset Z_0 \subset \overline{Z}_0$ by construction). Then $\overline{Z}  Z$ is a closed subset of $\overline{Z}$ that misses $\overline{Z}_0$. As $\overline{Z}$ is henselian along $\overline{Z}_0$, this means that $\overline{Z} = Z$, so $Y := \overline{Z}$ solves the problem.

$\begingroup$ That's a nice argument! Thanks! Since the latter part of the argument does not seem to use affineness, I guess it shows that at least in the Noetherian case one is done as long as one manages to lift every finite etale $Y_0 \rightarrow S_0$ to an etale (but not necessarily finite) $Y^{\prime} \rightarrow S$. $\endgroup$ – Kestutis Cesnavicius Jan 21 '14 at 14:49
The henselian pair question is in a paper of Gabber "Affine analog of the proper base change theorem" footnote on page 326. See also Lemma Tag 09ZS. Please also look further in Gabber's paper where he discusses combining the henselian pair picture with the proper base change theorem.