Let $(R,\mathfrak{m})$ be a DVR with fraction field $K$ and call $\hat{R}$ its completion. I would like to show that the henselization of $R$ consists of the element of $\hat{R}$ which are separable over $K$, trying to follow the path of this answer https://mathoverflow.net/questions/105381/henselization-and-completion.
I've got one problem with the proof that is showing that $B_F$ is etale over $R$ and in particular I do not get why $B_F$ should be finitely generated over $R$ .
My problem would be solved if the following thing is true:$\hat{R} \cap K^{sep}$ is exactly the set of elements $x \in \hat{R}$ that are integral and separable over $R$. In this case, $B_F$ would be the integral closure of $R$ in a finite separable extension and so it would be finite over $R$ (this shuld be a general fact about normal noetherian rings).I've tried to prove this claim, but I'm not getting anything.
My problem would be solved if the following thing is true:$\hat{R} \cap K^{sep}$ is exactly the set of elements $x \in \hat{R}$ that are integral and separable over $R$.
This is in general not true. Take for example $R = \mathbb{Z}_3 \cap \mathbb{Q} = \{\frac{a}{b}: a,b\in\mathbb{Z}, \gcd(a, b) = 1, 3\nmid b\}$. Then we have $K^{sep} = \bar{\mathbb{Q}}$ and $\hat{R} = \mathbb{Z}_3$.
Now there is an element $\alpha \in \hat{R}$ such that $3\alpha^2-2\alpha-2 = 0$ (the existence of $\alpha$ follows from Hensel's lemma). Thus $\alpha$ lies in the intersection $\hat{R} \cap K^{sep}$.
However, it is also clear that $\alpha$ is not integral over $R$.