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 $
Best Answer
The first red line follows because $|(z-a)f(z)|\le M|(z-a)|$ near $a$, where $M$ is the bound for $f(z)$. This clearly goes to zero as $z$ goes to $a$.
The coefficient $c_2$ is $h''(z)/2!$ evaluated at $a$. We see $h'(z)=(z-a)^2f'(z)+2(z-a)f(z)$ and $h''(z)=(z-a)^2f''(z)+2(z-a)f'(z)+2(z-a)f'(z)+2f(z)$. Plugging in $a$ and dividing by 2 gives the the value for $c_2$ above.