Let [t] denote the greatest integer not exceeding t. The number of points in the open interval (0,3) at which g(x)=[x^2] is discontinuous is

The function $[x^2]$ is locally constant except where $x^2$ crosses an integer, because $x^2$ is continuous and strictly increasing for $x>0$. So $g$ is discontinuous exactly at those $x>0$ for which $x^2$ is a positive integer. For $x\in(0,3)$ we have $x^2\in(0,9)$, so $x^2=1,2,\dots,8$, giving $x=1,\sqrt2,\sqrt3,2,\sqrt5,\sqrt6,\sqrt7,2\sqrt2$. That is $8$ points. The value $x^2=9$ corresponds to $x=3$, which is outside the open interval, and $x^2=0$ is not in the interval either; counting only the integers $x=1,2$ would give a wrong answer.