Let $D \subseteq \mathbb{C}$ be a domain and $\emptyset \neq U \subsetneq D$ open. I want to show the existence of a holomorphic function $f: U \rightarrow \mathbb{C}$ without holomorphic continuation to $D$.
Proof: Let $z_0 \in D \setminus U$ and consider
$$f: \mathbb{C} \setminus \{z_0\} \rightarrow \mathbb{C} \hspace{15pt} z \mapsto \frac{1}{z - z_0}$$
$f$ is holomorphic on $\mathbb{C} \setminus \{z_0\}$, especially on $U$. Now assume that there exists an holomorphic continuation $g$ of $f|_U$ on $D$. Since $f|_{D\setminus\{z_0\}}$ and $g|_{D\setminus\{z_0\}}$ are holomorphic on the domain $D\setminus\{z_0\}$ the identity theorem for holomorphic functions tells us $f|_{D\setminus\{z_0\}} \equiv g|_{D\setminus\{z_0\}}$, but there is no continuous and especially no holomorphic continuation of $f_{D\setminus\{z_0\}}$ to $D$ in contradiction to the existence of $g$.
Is my proof correct or can you come up with an easier proof?