Theorem
Let $\Omega\subseteq \mathbb{C}$ open , $ a\in\Omega,\ f\in H(\Omega\backslash \{a\})$ and there is $r>0$ with
$f$ is bounded on $C(a,r)\backslash \{a\}$ ($C(a,r)$ is the circle with origin $a$ and radius $r$), then $a$ is a removable singularity.
Proof
Let $h:\Omega\rightarrow \mathbb{C}$ be defined as: $ h(z)=\left\{\begin{array}{l} 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z=a \\ (z-a)^{2}f(z), \ z\in\Omega\backslash \{a\} \end{array}\right. $ Then we have: $ \lim_{z\rightarrow a}\frac{h(z)-h(a)}{z-a}=\lim_{z\rightarrow a}(z-a)f(z)=0\Rightarrow h'(a)=0. $
$\color{red}{\text{ Why is} \lim\limits_{z\rightarrow a}(z-a)f(z)=0?}$
So we have $h\in H(\Omega)$ and therefore $(h(a)=h'(a)=0)$. $ h(z)=\sum_{n=2}^{\infty}c_{n}(z-a)^{n}\ (z\in K(a,\ r)). $ Letting $f(a):=c_{2}$, it follows:
$\color{red}{\text{ Why do we have} f(a):=c_{2}?}$
$ f(z)=\sum_{n=0}^{\infty}c_{n+2}(z-a)^{n}\ (z\in K(a,\ r)), $ so $f\in H(\Omega).\ \square $