I do not know how to proof following question:
Let $D$ be a domain in $C$. ${{f}_{n}}$ be a sequence of holomorphic functions on $D$ that converges uniformly locally to a function non-constant $f$. Show that for each $z\in D$, there is a sequence ${{z}_{n}}\to z$ as ${{f}_{n}}({{z}_{n}})=f(z)$ for all sufficiently large $n$.
Actually, I know that I should use an assumption such as: ${{f}_{n}}({{z}_{n}})\ne f(z)$ Therefore, there is compact set that $\left\| {{f}_{{{n}_{k}}}}({{z}_{{{n}_{k}}}})-f(z) \right\|>\varepsilon $ Then I think I have to use Montel theorem, but I do not know how to show the contradiction.
Assume that $f$ is holomorphic in $D$, not constant, $f(z_0) = w_0$, and $f_n \to f$ locally uniformly in $D$.
The zeros of $f - w_0$ are isolated, so there exists an $r > 0$ such that $B(z_0, r) \subset D$ and $$ f(z) \ne w_0 \text{ for } 0 < |z - z_0| < r\, . \tag{*} $$
Let $(\varepsilon_k)$ be a sequence such that $0 < \varepsilon_k < r$ and $\varepsilon_k \to 0$. It follows from $(*)$ that $$ m_k := \min_{z \in \partial B(z_0, \varepsilon_k)} |f(z) - w_0| > 0 \, . $$ Since $f_n \to f$ uniformly on compact sets, there is an $n_k \in \mathbb N$ such that for all $z \in \partial B(z_0, \varepsilon_k)$ $$ |(f_{n_k}(z) - w_0) - (f(z) - w_0)| = |f_{n_k}(z) - f(z)| < m_k \le |f(z) - w_0| \, . $$
Now Rouché's theorem implies that $f_{n_k} - w_0$ and $f - w_0$ have the same number of zeros in $B(z_0, \varepsilon_k)$ (counting multiplicities). In particular, there is (at least) one $z_{n_k} \in B(z_0, \varepsilon_k)$ such that $f_{n_k}(z_{n_k}) = w_0$.
This sequence $(z_{n_k})$ satisfies $z_{n_k} \to z_0$ and $f_{n_k}(z_{n_k}) = w_0 = f(z_0)$.