Let $D\subseteq\mathbb C$ be open. Show that $(f_n)\stackrel{\text{comp}}\longrightarrow f$ implies $(f_n)\to f$ local uniform on D.
Notes:
1) A sequence $(f_n):D\to\mathbb C$ is said to converge compactly if $(f_n)\mid_V\to f$ local uniform for every compact subset $V\subseteq D$.
2) A sequence $(f_n):D\to\mathbb C$ is said to converge local uniform if for each point $z\in D$ there exists neighbourhood $V$ such that $(f_n\mid_V)\to f_V$ uniform.
Proof. Since $(f_n)\stackrel{\text{comp}}\longrightarrow f$ we get $(f_n)\mid_V\to f$ local uniform for each compact subset $V\subseteq D$. Now arbitrarily choose $z\in D$. Since $D$ is open, there is an open disk $K_\epsilon(z)\subset D$. Since $V:=\overline{K_{\epsilon/2}(z)}\subset D$ is compact, $(f_n)_V\to f_V$ must be local uniform.
Question: Is my proof correct?