Problem
Suppose $f$ being a holomorphic function and $f(0)=p$ as well as $f'(0)\neq0$. Proof that a positive number $r$ can be found such that the only solution to
$$f(0)=p\,\,,\,\,\,\,\,\,\,\,f:\mathbb{C}\rightarrow \mathbb{C}$$
that satisfies $\left | z \right | \leq r\,\,$ is $\,\,z=0$.
Attempt of Proof. $\,\,\,\,\,\,$ Given Laurent's Theorem which states that if $f$ is a holomorphic function on the annulus $\mathcal{A}(z_0,\,r_{in},\,r_{out})$, where $z_0$ is defined as its centre and $r_{out}$ width of outer radius and $r_{in}$ of inner, the Laurent series of $f$ on $\mathcal{A}$ converges to $f(z)$ on $\mathcal{A}$ :
$$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n \quad :\quad a_n=\int_{\gamma}\frac{f(\omega)}{(\omega-z_0)^{n+1}}d\omega$$
where $\gamma$ is any circle contained within $\mathcal{A}$. Apply $\mathcal{A}(z_0,0,r)$.
Consider two arbitrary points $z_i,\,z_j \in \mathcal{A}$ and obviously $z_i \neq z_j$ therefor. Assume that either of these points imply $f(z)=p\,$. Now, consider the Laurent series :
$$f(z_i)=\sum_{n=-\infty}^{\infty}a_{ni}(z_i-z_0)^n$$ $$f(z_j)=\sum_{n=-\infty}^{\infty}a_{nj}(z_j-z_0)^n.$$
Given that $a_{ni}=\int_{\gamma_i}\frac{f(\omega)}{(\omega-z_0)^{n+1}}d\omega\quad$ and $\quad a_{nj}=\int_{\gamma_j}\frac{f(\omega)}{(\omega-z_0)^{n+1}}d\omega$, $\,\,\,$applying $\gamma_i = \gamma_j$ implies $a_{ni}=a_{nj}:=\hat{a}_n$ and therefore
$$f(z_i)=\sum_{n=-\infty}^{\infty}\hat{a}_n(z_i-z_0)^n$$ $$f(z_j)=\sum_{n=-\infty}^{\infty}\hat{a}_n(z_j-z_0)^n$$.
Consider the difference $f(z_i)-f(z_j)$
$$f(z_i)-f(z_j)=\sum_{n=-\infty}^{\infty}\hat{a}_n\underbrace{\left[(z_i-z_0)^n-(z_j-z_0)^n\right]}_{\neq 0 \,\, \forall n \quad \because \quad z_i \neq z_j}\neq 0$$
and thus, if e.g. $f(z_i)=p$, then $f(z_i)-f(z_j)=p-f(z_j)\neq 0$ and $f(z_j)\neq p\,$. Apply $z_i=0$ and $z_0=0$ Q.E.D
Is this approach made correctly? Thankful for all input.
The proof cannot be correct, as it would show that generally $z_i \ne z_j$ implies that $f(z_i) \ne f(z_j)$, i.e. all holomorphic functions are injective, which is definitely wrong.
Essentially, you assume that for $z \ne 0$ $$ f(z) - f(0)= \sum_{n=1}^\infty a_n \underbrace{z^n}_{\ne 0} $$ must be different from zero. This is a wrong conclusion, a simple counterexample is $f(z) = \sin(z)$.
But it is much simpler: Assume that such a positive number $r$ does not exist. Then for arbitrary $r > 0$ there is some $z \ne 0$ with $|z| < r$ and $f(z) = f(0)$. In particular, there is a sequence $(z_n)$ such that $\lim_{n \to \infty} z_n = 0$ and $$ \forall n: z_n \ne 0 \,, \, f(z_n) = f(0) \, . $$ It follows that $$ f'(0) = \lim_{n \to \infty} \frac{f(z_n) - f(0)}{z_n - 0} = 0 \, , $$ contrary to the assumption.