Algebraic closures are henselian? Let $(K,v)$ be a nonarchimedean valued field and $(\widehat{K},\widehat{v})$ be its completion.
Let $o$ and $\widehat{o}$ be the valuation rings of $K$ and $\widehat{K}$.
Let $K_v$ be the separable closure of $K$ in $\widehat{K}$. Let $o_v:=K_v\cap \widehat{o}$.
Let $f(x)=a_nx^n+\cdots+a_0\in o_v[x]$ be a primitive polynomial, i.e. $\max\{|a_n|,\cdots,|a_0|\}=1$.
Let $g,h\in \widehat{K}[x]$ be polynomials with $f=gh$ and assume that the highest coefficient of $g$ is a unit in $o_v$.
If $K_v$ is algebraically closed in $\widehat{K}$, then all coefficients of $g$ and $h$ are 
in $o_v$?

I found this question while searching for clarification on that same section of the book. Have you found a satisfying explanation?
By the way from my understanding, the argument is that the coefficients are polynomial functions of the roots of $f$, which are obviously algebraic over $\mathcal{O}_v$ by definition of $f$, and so the coefficients are algebraic as well (because algebraic numbers form a field).
For $g$, after multiplying by a suitable constant (and dividing $h$ by the same constant) such that the highest degree coefficient of $g$ is in $\mathcal{O}_v^*$, we can then use Vieta's formulas to argue that the coefficients equal up to units the elementary symmetric polynomials in the roots of $g$ (which are roots of $f$). Hence they are also algebraic. Then $h = f/g$ so its coefficients will also be algebraic by the division algorithm.
So $f$ factors over $K_v$ (because it was assumed to be algebraically closed, i.e. $K_v = \overline K$), and then by Gauss' lemma $f$ also factors over the ring of integers $\mathcal{O}_v$.
What I am wondering is, wouldn't this argument imply that not only the separable closure of $K$, but also the algebraic closure of $K$, should be henselian?
