The question is to show, given a local domain $A$, with maximal ideal $(\pi)$, $B$ a domain containing $A$, $S=A-(\pi)$. If $S^{-1}B$ is integrally closed, and $B/\pi B$ reduced, then $B$ is integrally closed.
My proof is to take $x\in\mathrm{Quot}(B)$ integral over $B$, this will be integral over $S^{-1}B$, so will be contained in $S^{-1}B$, thus there exists $s\in A-(\pi)$ with $sx\in B$.
But in $A$, $s$ is a unit, so is a unit in $B$, so $x\in B$, and we are done.
This argument doesn't use that the maximal ideal is principle, or the reduced hypothesis, so I suspect there's an error somewhere, but I cant see where.