If a continuous path $\Xi$ in $A\subset\mathbb{R}^2$ starts and end on $\partial(A)$, show that $A-\Xi$ is disconnected If a continuous path $\Xi$ in a closed and bounded subset $A\subset\mathbb{R}^2$ starts and end on $\partial(A)$, show that $A-\Xi$ is disconnected
To make things formal let $T=[0,1]$ and say that $\Xi$ is a pair of continuous functions $x \colon T \to \mathbb{R}$,$y \colon T \to \mathbb{R}$ such that
$\forall\,t\;(x(t),y(t))\in A$ 
$(x(0),y(0))\in \partial A$
$(x(1),y(1))\in \partial A$
$\exists\;t\;s.t.\;(x(t),y(t))\in int(A)$. 
Show that $A-\Xi$ is disconnected. This is obvious from any illustration, but finding a rigorous proof eludes me. If instead we are working in the unit square for example and the path $\Xi$ is instead the graph of a continuous function $f:[0,1] \to [0,1]$ then I can do it.
Define the sets $U=${$(x,y):f(x)>y$}$,\;L= ${$(x,y):f(x)<y$}$ $ U and L are open because f is continuous. They are both nonempty because the graph of f has a point in the interior. and they obviously partition the space $[0,1]^2-\Xi$ so we are done.
any help or references in the general case would be heavily appreciated.
Thanks

I don't think this is true. Take $A = [0,1]\times [0,2]$ and let $(x,y) : T \to [0,1]^2$ be surjective, i.e.a space filling curve, which we can easily take to start and end in $(0,0) \in \partial A$.
In this case, $A \setminus \Xi = [0,1]\times (1,2]$ is connected.
