Connectedness of a certain subset of the plane Let $U$ be an open and connected subspace of the Euclidean plane $\mathbb{R}^2$ and $A\subseteq U$ a subspace which is homeomorphic to the closed unit interval. Is $U\setminus A$ necessarily connected?

Every subset $A$ of $\mathbb{R}^{2}$ homeomorphic to $[0,1]$ is "tame", that
is, there is a self-homeomorphism of the plane $\varphi$, such that
$\varphi(A)=[0,1]$ (citation needed :)). Then it follows that $A$ may be
represented as an intersection $A=\cap_{i=1}^{\infty}D_{i}$ of a decreasing
sequence of (closed) topological disks, so we have $D_{i}\subset U$ for some
$i$ (sufficiently large). Now, to show that $U\backslash A$ is connected, let
$x,y\in U\backslash A$ and $\gamma$ be a topological arc in $U$ connecting $x$ and $y$;
then the set $L=(\gamma\backslash D_{i}^{0})\cup\partial D_{i}$ is connected.
To prove this, let $L=L_{1}\cup L_{2}$ where $L_{1}$, $L_{2}$ are disjoint
open subsets of $L$, then since $\partial D_{i}$ is connected, we may suppose
that $\partial D_{i}\subset L_{1}$, but then each component $K$ of
$\gamma\backslash D_{i}^{0}$ is also contained in $L_{1}$ ($K$ is intersecting
$\partial D_{i}$ as$\ \gamma$ is connected!); thus $L=L_{1}$, so $L$ is
connected. Now, since $L\subset U\backslash A$, it follows that any two points
of $U\backslash A$ are contained in a connected set and therefore $U\backslash
A$ is connected as well.
p.s. [Here $D_{i}^{0}$ is the interior of $D_{i}$.]
