Let $\mathcal O (\Omega)$ denote the set of all holomorphic function on a domain $\Omega\subset \mathbb C$. Define the holomorphic hull of a compact subset $K$ of $\Omega$ as $$ \hat K_{\mathcal{O}(\Omega)}=\{z\in\Omega\;;\;\lvert f(z)\rvert\leq\max_{z\in K}\lvert f(z)\rvert\ \text{ for all } f\in\mathcal O(\Omega)\}. $$ The question is the following. Let $M>0, \varepsilon>0$ and $p\in\Omega-\hat K_{\mathcal O(\Omega)}$. Prove the existence of a function $f\in\mathcal O(\Omega)$ such that $\max_{z\in K}\lvert f(z)\rvert<\varepsilon$ and $\lvert f(p)\rvert > M$.
My humble attempt: If $p\in\Omega-\hat K_{\mathcal O(\Omega)}$, $\exists f\in \mathcal O(\Omega)$ such that $\max_{z\in K}\lvert f(z)\rvert<\lvert f(p)\rvert$. Defining $g(z):=\frac{M}{\max_{z\in K}\lvert f(z)\rvert}f(z)$, I get $\lvert g(p)\rvert > M$.
But I don't see how I can prove the other inequality. Invoking Runge's theorem seems necessary, but I don't know exactly how to achieve this.
Figured it out thanks to @totoro. Mergelyan's theorem states that if $K$ is compact in $\mathbb C$ and $\mathbb C-K$ is connected, every continuous function $f:K\rightarrow\mathbb C$ such that $f\rvert_{\operatorname{int}(K)}$ is holomorphic can be approximated uniformly on $K$ with polynomials.
Denote $K_0=\hat K_{\mathcal O(K)}$. Let $\varphi(z):=M\frac{\operatorname d(z,K_0)}{\operatorname d(z, K_0)+\operatorname d(z,p)}$. It can be shown $\operatorname d(\cdot,K_0)\colon \mathbb C\rightarrow\mathbb R$ is uniformly continuous (and the metric is continuous too, of course) and the denominator is nowhere zero. There holds $f\rvert_{K_0}\equiv0\equiv f\rvert_{\operatorname {int}(K_0)}$ and $f(p)=M$. By Mergelyan's theorem, there exists a holomorphic polynomial $g$ such that $\lvert g - \varphi\rvert_{K_0\cup\{p\}}<\frac\varepsilon 2$, which implies $|g|_{K_0}<\frac \varepsilon 2$ and $|g(p)-M|<\frac \varepsilon 2$. Now $f(z)=g(z)+\frac \varepsilon 2$ is the desired function.