Assume that $F=k$ and let $p=char k$.
If $K/k$ is inseparable then there exists $\alpha\in K$ such that min. poly of $\alpha$ over $k$ is equal $g(X^{p^n})$ with $g$ irreducible separable. Since $R$ is a valuation ring, either $\alpha$ or $\alpha^{-1}$ is in $R$. We may assume that $\alpha\in R$ by symmetry. Let $\bar{\alpha}$ be the image of $\alpha$ in $k$. Then $\bar{\alpha}^{p^n}$ is a root of $g$ in $k$, hence $g(X)=X-\bar{\alpha}^{p^n}$, which implies $(\alpha-\bar{\alpha})^{p^n}=0$ in $R$. Since $R$ is a domain $\alpha=\bar{\alpha}\in k$.
Assume that $F=k$ and let $p=char k$.
If $K/k$ is inseparable then there exists $\alpha\in K$ such that min. poly of $\alpha$ over $k$ is equal $g(X^{p^n})$ with $g$ irreducible separable. Since $R$ is a valuation ring, either $\alpha$ or $\alpha^{-1}$ is in $R$. We may assume that $\alpha\in R$ by symmetry. Let $\bar{\alpha}$ be the image of $\alpha$ in $k$. Then $\bar{\alpha}^{p^n}$ is a root of $g$ in $k$, hence $g(X)=X-\bar{\alpha}^{p^n}$, which implies $(\alpha-\bar{\alpha})^{p^n}=0$ in $R$. Since $R$ is a domain $\alpha=\bar{\alpha}\in k$.