"weak" Henselian property I encountered following thread from MO treating the structure $X(k)$ of $k$-valued points of a separated algebraic space $X$ of finite type over $k$.
I have a question about an aspect from Laurent Moret-Bailly's amazing answer. let me quote: 
"...Assume now that $k$ satisfies the following "weak henselian" property: 
(WH) Every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$. ..."
Question: why the condition (WH) that every étale morphism $Y\to X$ gives rise to an open map $Y(k)\to X(k)$ can here be called "weak henselian property"? why it is a "weak" form of henselian property?
Recall: a ring $R$ is henselian if it satisfies this equivalent properties. in a more general context take a look to Milne's book "Etale Cohology" Thm 4.2 (page 32).
My question is still is which sense (WH) can be considered as "weaker" form of these listed "classical" henselian properties from above?
Thank you in advance!
 A: Let me make (almost) everything explicit.
A valued field $(k,v)$ (whose valuation ring and residue field I will respectively denote by $\mathcal{O}$ and $\kappa$) is said to be henselian whenever it verifies Hensel's lemma, the statement of which is exactly the same as the one you might know for $\mathbf{Q}_p$:

If $f\in\mathcal{O}\left[X\right]$ is monic then every root of
  multiplicity one of $\overline{f}\in\kappa\left[X\right]$ (the residue
  of $f$ in kappa) lifts uniquely to a root of $f$ in $\mathcal{O}$.

As you mentioned, this is equivalent to saying that the local ring $\mathcal{O}$ is henselian and the link you give from the Stakcs Project gives several equivalent conditions.
Nevertheless, the fact that $\mathcal{O}$ is a valuation ring gives you additional properties (I mean other than the one from the Stacks Project's link you gave) equivalent to Hensel's lemma, which you can find in Chapter 9 of Kuhlmann's book on Valuation theory. I will use the following property which is equivalent to Hensel's lemma (the caption is excerpt from Kuhlmann's book) and which is a form of the "implicit function theorem" that KReiser seems to mention in his comment:

As an exercise, you can deduce the following variation of the prequel:

Corollary: If the $f_i$'s are being taken over $k$ (not only over $\mathcal{O}$)
  and $z=(x,y)$ is a common zero of the $f_i$'s in $k^{m+n}$ such that
  $J(z)\neq 0$, then there exists $\alpha>0$ such that for all
  $x'=(x'_1,\dots,x'_m)\in x+\mathcal{O}^m$ verifying
  $v(x'_i-x_i)>2\alpha$ (for $1\leq i\leq m$) there exists a unique
  tuple $y'=(y'_1,\dots,y'_n)\in y+\mathcal{O}^n$ such that $(x',y')$ is
  a common zero of the $f_i$'s and $v(y'_i-y_i)\geq v(x'_i-x_i)-\alpha$.

Hence the following "implicit function theorem":

Implicit function theorem: If the $f_i$'s are taken over $k$ and
  $z=(x,y)\in k^{m+n}$ is a common root of the $f_i$'s such that
  $J(z)\neq0$, then there exists open neighbourhoods $U$ of $x$, $V$ of
  $y$ and $\phi:U\rightarrow V$ continuous such that for all $(x',y')\in U\times V$ one has $f(x',y')=0$ if and only if $y=\phi(x)$.

In the thread you pointed to, Laurent Moret-Bailly claims the following:

Claim 1: if $(k,v)$ is a henselian valued field then for any $f:Y\rightarrow X$ étale morphism of $k$-varieties (here "variety" stands for locally of finite type), the induced continuous map $Y(k)\rightarrow X(k)$ is open.

Actually, one can do better:

Claim 2: if $(k,v)$ is a henselian valued field then for any $f:Y\rightarrow X$ étale morphism of $k$-varieties (here "variety" stands for locally of finite type), the induced continuous map $Y(k)\rightarrow X(k)$ is a local homeomorphism.

To answer your first question, I will prove Claim 2. Actually, I do not know what your background is, but you first might be wondering what the topology on $X(k)$ is, whenever $X$ is a $k$-variety and $k$ is a topological field: to say it very quickly, first assume that X is affine with ring of functions $k\left[X_1,\dots,X_n\right]/(f_1,\dots,f_r)$ so that $X(k)$ is identified to the zero locus of $(f_1,\dots,f_r)$ in $k^n$ and endow this zero locus with the topology induced by that of $k^n$; if $X$ is not affine, one verifies that the topologies may be glued.

Proof of Claim 2: Let $(k,v)$ be a henselian field and $f:Y\rightarrow X$ an étale morphism
  of $k$ varieties. Clearly, one may assume that $Y$ and $X$ are affine:
  since $f$ is étale, write $X=\mathrm{Spec}(A)$ where
  $A=k\left[y_1,\dots,y_r\right]/(g_1,\dots,g_s)$ and
  $Y=\mathrm{Spec}(B)$ where
  $B=A\left[x_1,\dots,x_n\right]/(f_1,\dots,f_n)=k\left[x_1,\dots,x_n,y_1,\dots,y_r\right]/(f_1,\dots,f_n,g_1,\dots,g_s)$
  and $J(x_1,\dots,x_n,y_1,\dots,y_r)=\mathrm{det}\left(\frac{\partial
f_i}{\partial x_j}\right)_{1\leq i,j\leq n}\in B^\times$.
Then, $Y(k)\subseteq k^{n+r}$ is the zero locus of
  $(f_1,\dots,f_n,g_1,\dots,g_s)$ and $X(k)\subseteq k^r$ is the zero
  locus of $(g_1,\dots,g_s)$ and the map $f(k):Y(k)\rightarrow X(k)$
  maps a tuple $(x_1,\dots,x_n,y_1,\dots,y_r)$ to the tuple
  $(y_1,\dots,y_r)$. Let $\xi=(x_1,\dots,x_n,y_1,\dots,y_r)\in Y(k)$:
  since $\xi$ is a common zero of the $f_i$'s and $J(\xi)\neq 0$, the
  implicit function theorem gives an open neighbourhood $V$ of
  $(y_1,\dots,y_r)$ in $X(k)$ and an open neighbouhrood $U$ of
  $(x_1,\dots,x_n)$ in $Y(k)$ such that $f(k)$ induces a homeomorphism
  $U\rightarrow V$. Thus, $f(k)$ is a local homeomorphism which proves
  Claim 2.

As for your second question, it is not very clear. Actually, if I understood well, you would like to avoid valuations and deal with $k$ as the fraction field of a henselian local ring $R$ (which is not necesarily a valuation ring): but then, what is the topology you give to $k$? The only one I could think about is the $\mathcal{m}_R$-adic one but then you would not retrieve the case of a henselian valued field, since there exist henselian valued fields whose valuation group has an uncountable cofinality. 
