Old question: Let $R$ be a UFD, and $Q(R)$ its field of fractions.
Let $S$ be an overring of $R$, namely, a ring such that $R \subseteq S \subseteq Q(R)$.
Question: Is it true that $S$ is a UFD?
Notice that $R=k[x^2]$, $S=k[x^2,x^3]$ is not a counterexample, since $S$ is not contained in the field of fractions of $R$, $Q(R)=k(x^2)$.
Please see this similar question: Here we require that $S \subseteq Q(R)$ (and remove the requirement for simplicity); the answer there presents rings $R \subseteq S$ with $S \not\subseteq Q(R)$.
Thank you very much!
New question: After receiving several comments, including the following counterexample $k[x,y] \subset k[x,y,y/x] \subset k(x,y)$, I would like to change my question to the following one:
New question: Is $S$ normal, namely, integrally closed in its field of fractions $Q(S)$?
This and this paper are relevant. If I am not wrong, $(k[x],k(x))$ is a normal pair, namely, every ring $k[x] \subseteq S \subseteq k(x)$ is integrally closed in $k(x)$, so for $R=k[x]$ the new question has a positive answer.
More generally, if I am not wrong, if $R$ is a Dedekind domain, then $(R,Q(R))$ is a normal pair, hence answers the new question positively.
What about $R=k[x,y]$? Is $(k[x,y],k(x,y))$ a normal pair?
Let $R=K[x,y]$ and $S=K[x,y,\frac{y^2}{x^2}]$. Since $\frac yx$ is integral over $S$ and does not belong to $S$ it follows that $S$ is not integrally closed.