Suppose $p_1, \cdots p_n$ are points on the compact Riemann surface $X$ and $X^{\prime}=X\setminus \{p_1,\cdots, p_n\}$. Suppose \begin{equation*} f:X^{\prime} \longrightarrow \mathbb{C} \end{equation*} is a non-constant holomorphic function. Show that the image of $f$ becomes arbitrarily close to every $c \in \mathbb{C}$.
Here is my argument. According to the question we are asked to show that $\overline{f(X^{\prime})}=\mathbb{C}$, i.e., the image of $X^{\prime}$ under $f$ is dense in $\mathbb{C}.$ For a contradiction assume that $f(X^{\prime})$ is not dense in $\mathbb{C}$. Then this means that there exists an open subset(a ball of radius $r$ centered at $c$) in $\mathbb{C}$, say $B_{r}(c)$ so that $B_{r}(c) \cap f(X^{\prime})=\emptyset $. This is to say that $f(z)\neq c$ for some $z \in X^{\prime}$. If this is the case then one can define a function $g(z)=\frac{1}{f(z)-c}$ which is holomorphic on $X^{\prime}$. Since $|f(z)-c|>r$, one has $|g(z)|< \frac{1}{r}$. Thus the function $g$ is holomorphic and bounded. But I could not complete the proof. I could not even use the condition that the points $p_i$ are removed. How do I need to proceed? Thanks in advance.