Let $U$ be the open unit disk in $\mathbb{C}$ and let $A$ be an open non-empty subset of $U$ that is not dense in $U.$
Under these conditions, why is it then true that there exists $\alpha \in A,$ $\beta \in U - \bar{A}$ such that $$2 |\beta-\alpha| < 1-|\beta|?$$ This should be straightforward, but I can not see why at the moment.
Below, closures and boundaries are taken with respect to $U.$
Because $U\setminus \bar A$ is nonempty, $\partial \bar A$ is nonempty, otherwise $A = \bar A,$ contradicting the connectivity of $U.$ Thus there exists $c\in \partial \bar A = \partial (U\setminus \bar A).$ Hence there exist sequences $a_n$ in $\bar A$ and $b_n$ in $U\setminus \bar A$ converging to $c.$ We can then find $a_n'\in A$ converging to $c.$ It follows that $|a_n'-b_n|\to 0.$ Note that $1-|b_n|\to 1-|c|>0.$ This implies
$$|a_n'-b_n| < (1-|b_n|)/2$$
for large $n.$ This gives the desired result.