Topological disk doesn't separate the plane We know the Jordan arc theorem that says that if $\gamma \subset \mathbb{R}^2$ is a homeomorphic image of the interval $[0,1]$, then the set  $\mathbb{R}^2 \backslash \gamma$ is connected.
How to prove or where one can find a referece for the fact that if $K \subset \mathbb{R}^2$ is a homeomorphic image of the closed unit disk $B_1 \subset \mathbb{R}^2$, then the set  $\mathbb{R}^2 \backslash K$ is connected.
EDIT:  If we use the Jordan curve theorem and pick a homeomorphism $f:B_1 \rightarrow f(B_1)=K$, then is it clear that the set $f(int(B_1))$ is open and therefore a connected and bounded component?
 A: Bredon's guideline to the proof of the Jordan curve theorem includes the following theorem (which he attributes to Alexander):

Theorem: Let $n$ be fixed. Suppose that $Y$ is a compact space with the property that $\widetilde{H}_*(S^n-f(Y))=0$ for every embedding $f:Y \to S^n$. Then $I \times Y$ also has this property.

The argument of the proof of the above theorem is essentially Mayer-Vietoris together with a direct limit argument. As a corollary through induction, one gets:

Corollary: If $f: D^r \to S^n$ is an embedding, then $\widetilde{H}_*(S^n-f(D^r))=0.$ 

Since $S^n-f(D^r)$ is open in the sphere, it follows that it is connected. Of course, the case for $\mathbb{R}^n$ follows simply because if you have an embedding to $\mathbb{R}^n$, you have an embedding to the sphere. And taking away a point from an open set of the sphere (with $n>1$) will not change its connectedness.
A: The fastest way to prove this is via (a special case of) the Alexander duality: For an arbitrary compact subset $A$ of $S^n$,
$$
\tilde{H}_0(S^n -A)\cong H^{n-1}(A)
$$
where the cohomology of $A$ is Chech. If $A$ is contractible, e.g. homeomorphic to a disk, then $H^{n-1}(A)=0$ and, hence, $\tilde{H}_0(S^n -A)=0$, i.e. $H_0(S^n -A)\cong {\mathbb Z}$ and, thus, $S^n -A$ is connected. In your case, $A=K\cong D^2$ is a compact subset of $R^2$, hence, it does not separate $S^2$, hence, does not separate $R^2$ either. When $n=2$ and $A\cong D^2$ one can give other arguments, but they are special to the 2-dimensional case and, in my mind, obscure the problem. If you want to avoid the Alexander duality, take a look at Massey's book
"Algebraic Topology: An Introduction."
He proves the Jourdan separation theorem in $S^n$ by computing first $H_0(S^n -A)$, where $A$ is homeomorphic to a closed disk $D^k$. In Massey's proof $A\cong D^{n-1}$ but his argument works for any $k$.  
