Path connectedness in unit disk Let $D\subset \mathbb{R}^2$ be the closed unit disk, boundary of $D$ is $\mathbb{S}^1$, let $a,b\in \mathbb{S}^1$ are two distinct points, $A,B\subset D$ are two disjoint closed sets with $A\cap \mathbb{S}^1=\{a\}$ and $B\cap \mathbb{S}^1=\{b\}$. 
My problem is that given any distinct $x,y \in \mathbb{S}^1-\{a,b\}$, could $x,y$ be connected by a continuous path which lies in  $D-A\cup B$? This means that there exists a continuous $f:[0,1]\rightarrow D-A\cup B$ such that $f(0)=x$ and $f(1)=y$.
 A: At first let me introduce some simple notation. For a closed bounded set $A \subset \mathbb{R}^2$ I will denote by $\tilde{A}$ the union of $A$ and all bounded components of $\mathbb{R}^2 \setminus A$. Alternatively $\mathbb{R}^2 \setminus \tilde{A}$ is the unbounded component of $\mathbb{R}^2 \setminus A$. I will make some simple observations about this thing.


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*if $A \subset D$ then $\tilde{A} \subset D$ and $A \bigcap S^1 = \tilde{A} \bigcap S^1$.

*if $A \subset D$ and $A \bigcap S^1 = \{a\}$ then $D \setminus \tilde{A}$ is connected.

*if $A$ and $B$ are disjoint then you can introduce $A_0 = A \bigcap \tilde{B}$ and $B_0 = B \bigcap \tilde{A}$. In this case $A_0$ and $B_0$ are the components of $A$ and $B$ respectively and for $A' = A \setminus A_0$ and $B' = B \setminus B_0$ we have $\tilde{A} \bigcup \tilde{B} = \tilde{A'} \bigcup \tilde{B'}$ and $\tilde{A'} \bigcap \tilde{B'} = \emptyset$
From these observations it is easy to derive that all we need is to prove that $D \setminus (\tilde{A'} \bigcup \tilde{B'})$ is connected (under your assumptions).
Now we can proceed by proving one simple fact. If $X$ is locally path connected and simply connected and $F,G \subset X$ are two disjoint closed sets such that $X \setminus F$ and $X \setminus G$ are connected then $X \setminus(F \bigcup G)$ is also (path) connected.
For $U = X \setminus G$ and $V = X \setminus F$ we apply Mayer-Vietoris sequence: $\dots \rightarrow \tilde{H}_1(X) \rightarrow \tilde{H}_0(U \bigcap V) \rightarrow \tilde{H}_0(U) \oplus \tilde{H}_0(V) \rightarrow \tilde{H}_0(X) \rightarrow 0$ is exact. But all terms except $\tilde{H}_0(U \bigcap V)$ are trivial and therefore $U \bigcap V$ is connected. But $U \bigcap V = X \setminus(F \bigcup G)$.
A: No.  Let $A = \{a\}$ and $B$ a circle inside $D$ tangent to S$^1$ at $b$.
Let $x$ be the center of the circle and $y$ a point on S$^1$ other than $a$ or $b$.
