In my complex analysis course the professor used the folowing fact: If $U \subset \mathbb{C}$ is open and $f: U \to \mathbb{C}$ is a holomorphic function (we defined this as $f$ being differentiable and its derivative being continuous) then the function $g: U^2 \to \mathbb{C}$
$ g(z, c) = \begin{cases} \frac{f(z) - f(c)}{z - c} & z \neq c \\ f'(c) & z = c \end{cases} $
is continuous.
I can prove this by using the fact that $f$ is equal to a power series but I would like to see a proof that doesn't use this since we haven't proved it in my course.
Expressing $f(z)$, $f(c)$ and $f'(c)$ by the Cauchy integral formula, you find $$ g(z,c)=\frac1{2\pi i}\oint_\gamma \frac{f(\zeta)}{(\zeta-z)(\zeta-c)} $$ where $\gamma$ is a simple closed curve surrounding $z$ and $c$. This is clearly continuous.