Consider an algebraically closed field $k$, a finite field extension $K$ of $k(T)$, the integral closure $A$ of $k[T]$ in $K$, and the integral closure $A'$ of $k[T^{1}]$ in $K$. Does it follow that $A \cap A' = k$?

Yes. Each element of $K$ satisfies a unique irreducible monic polynomial over $k(T)$. It is integral over $k[T]$ if and only if the coefficients lie in $k[T]$ and integral over $k[T^{1}]$ if and only if the coefficients lie in $k[T^{1}]$. If it is integral over both, the coefficients lie in $k[T] \cap k[T^{1}]=k$. Because $k$ is algebraically closed, an irreducible polynomial with coefficients in $k$ is linear. 

