In the complex plane $\ \mathbb C,\ $ consider six vertices (they belong to the unit circle $\ S:=D\setminus U\ $ that is the boundary of $D$):
$$ a_1:= 1\qquad\text{and}\qquad A_1=-a_1; $$ $$ a_2:= \frac{-1-i}{\sqrt 2}\qquad\text{and}\qquad A_2=-a_2; $$ $$ a_3:= i\qquad\text{and}\qquad A_3=-a_3; $$
Then consider the nine (curved or straight) edges that connect vertices $\ a_k\ $ to vertices $\ A_m,\ $ where six of those edges are the just obtained arcs of $\ S,\ $ and two of them are the given arcs that connect $\ a_1\ $ to $\ A_1,\ $ and $\ a_3\ $ to $\ A_3,\ $ and the ninth edge is the straight diameter interval that connects $\ a_2\ $ and $\ A_2.\ $
If the mentioned two arcs are disjoint then every pair consisting of the different arcs of the said nine arcs have disjoint interiors. Thus our graph of six vertices and nine edges would be isomorphic to the second Kuratowski's graph $\ K_{3,3}.\ $ This however would mean that our graph is not planar -- the two special edges have to intersect.