Let $R$ be a domain and $K$ be its quotient field. Is it true that if every ring $W$ with $R \subsetneq W \subset K$ is a valuation domain, then $R$ is a valuation domain?
Here, a domain $R$ is said to be a valuation domain if for any $a,b \in R$ either $a$ divides $b$ or $b $ divides $a$.
I think this is true and I tried to prove this by assuming that $R$ is NOT a valuation ring to get a contradiction. But I couldn't get the desired conclusion. Any help will be appreciated.
The statement is not true. A counterexample is $$ R=\left\{\frac{a}{b}\in\mathbb{Q}:(6,b)=1\right\}. $$ Then $R$ is not a valuation domain because $2\nmid 3$ and $3\nmid 2$. In fact, $R$ is a $1$-dimensional semi-local ring, with three prime ideals: $(0)$, $(2)$, and $(3)$. The only rings $W$ with $R\subsetneq W\subset K=\mathbb{Q}$ are $W=\mathbb{Z}_{(2)}$, $W=\mathbb{Z}_{(3)}$, and $W=\mathbb{Q}$, and all of these are valuation domains.