$A = (0,1]\cup[1,3] = (0,3]$. The interior $A^0$ consists of all interior points of $A$. The point $0$ is not in $A$; the point $3$ is a boundary point (right endpoint). Therefore $A^0 = (0,3)$. But $A = (0,3]$, so $A^0 = (0,3) = (0,1)\cup(1,3)\cup\{1\}$... The open core of $(0,3]$ is $(0,3)$.