I was reading Priestley's "Introduction to Complex Analysis", In Chapter 11, particularly, Examples 11.8, second example
$f(z) = \csc^2 z$ then $\int_{\gamma} f(z)=0$ where $\gamma$ is the unit disc centered at origin.
They used Fundamental Theorem of Calculus, in that $f = F'$ should be continuous in $\gamma^*$ and $F$ should be defined in an open set containing $\gamma^*$ then $\int_{\gamma} f(z)=0$ when $\gamma$ is a closed path.
Here $f(z)=F'(z)= \csc^2(z)$ which is continuous in $\gamma^*$ but $F(z)= \cot z$ is not defined in the open set containing $\gamma^*$ because $\cot z$ is not defined at $z=0$ which lies inside our $\gamma$.
So, how did they use the theorem? Please tell me if I am right or if I am wrong.
The problem is with how you chose your open set: you have opted to include $0$ in it, where indeed $\cot$ explodes. The idea is to choose the open set to contain $\gamma ^*$, on the one hand, and to avoid all the singularities and poles of $\cot$, on the other.
Notice that $\cot$ has poles at $\{ k \pi \mid k \in \Bbb Z \}$ (and no singularities). The closest points to $\gamma ^*$ from this set are $\pm \pi$ and $0$, therefore we want an open set avoiding these. Then the circular crown
$$\{ z \in \Bbb C \mid a < |z| < b \}$$
is good, for arbitrary $a \in [0,1)$ and $b \in (1, \pi]$.
There are infinitely many possibilities, the one above is just the easiest to write explicitly.