I've managed [with help from the wonderful people lurking on this site :)] to prove that for an integral domain $A$, if $A$ is integrally closed, then $S^{-1}A$ is integrally closed for all multiplicatively closed subsets $S$ of $A$.
The problem that I want to apply this to is:
$A$ is integrally closed if and only if $A_{P}$ is integrally closed for all maximal ideals $P$ of $A$.
So far:
$(\Rightarrow)$ If $P$ is a maximal ideal of $A$, then $A\setminus P$ is multiplicatively closed, because maximal ideals are prime. So by the result mentioned above, $A_{P} = (A \setminus P)^{-1}A$ is integrally closed.
Update: Attempting to prove the contrapositive.
Assume there is some $\frac{r}{s}$ in $F$ (the field of fractions of $A$) such that $\frac{r}{s}$ is integral over $A$ but not in $A$. I feel like I haven't done hardly anything at all with your hint, but I suppose that I would want to somehow get a maximal ideal of $A$ from this $\frac{r}{s}$ such that $A_{P}$ is not maximal, thus proving the converse.
Best Answer
You will need the fact that for any domain $A$, $\bigcap_{\mathfrak{m} \in \operatorname{MaxSpec} A} A_{\mathfrak{m}} = A$. If you are not familiar with this, see e.g. $\S 7.7$ of my commmutative algebra notes.
Armed with this fact, the proof is virtually immediate: let $K$ be the fraction field of $A$, and suppose that an element $x \in K$ is integral over $A$. It must then be integral over the larger ring $A_{\mathfrak{m}}$ for each $\mathfrak{m} \in \operatorname{MaxSpec} A$, so...?
(Note that this approach does not involve proving the contrapositive, although presumably one can make that work as well...)