A sequence has no limit point in unit disk, then $D_1 \setminus \{a_n\}$ is open. I need to show the if $\{a_n\} \subset \Bbb{C}$ is a sequence with no limit point in the unit disk, show unit disk minus $\{a_n\}$ is open,
So I want to go about it by contradiction. So if $D_1 \setminus \{a_n\}$ is not open, then it contains a boundary point. Say some $z_0$ such that $B_\epsilon(z_0)$ contains points both inside and outside of $D_1 \setminus \{a_n\}$. Any hints greatly appreciated.

Given the sequence $(a_n)\subset \mathbb{C} $ has no limit point in $D_1=D(0, 1) $
Let $A=\{x_n:n\in\Bbb{N}\}$
Claim : $D_1\setminus A$ is open.
Proof: Assume the contrary that $D_1\setminus A$ is not open.
Then $\exists x_0\in D_1\setminus A$ such that $\forall r>0, D(x_0, r) \not\subset D_1\setminus A $
Hence $\forall n\in\Bbb{N} ,\exists (x_n) \subset A$ such that $x_n\in D(x_0, \frac{1}{n}) $.
Since $(x_n)\subset A$ and $(x_n)\to x_0\in D_1$ implies $x_0\in D_1$ is a limit point of $A$.
Contradiction.
