Let $D$ be a bounded domain (open connected) in $ \mathbb C$ and assume that complement of $D$ is connected.Then show that $\partial D$ is connected I am trying to prove the following famous result in Point Set Topology. 

Let $D$ be a bounded domain (open connected) in $ \mathbb C$ and assume that complement of $D$ is connected. Then show that  $\partial D$ is connected.

I know there are proofs of this result using Fundamental Group and Algebraic Topology but I don't know much Algebraic Topology so I am trying to prove this result using Elementary Topology only.
I am planning to use following basic result of Point Set Topology:

Let $(K_i)_{i\in I}$ be an indexed family of decreasing, connected, compact sets in a Topological space X. Then $B = \bigcap_i K_i$ is connected.

I defined  $K_n$= {$ z \in D : dist (z, \partial D) \leq \frac{1}{n}$}. Now as distance is a continuous function therefore each $K_n$ is compact and also $K_n$ is a decreasing sequence. Furthermore, $ \bigcap K_i$=$\partial D$. So I think now its enough to show that each $K_i$ is connected. I personally believe that each $K_i$ is path connected, but I am unable to prove this. Feel free to scold me If I am doing something wrong here. Please help with my idea or give some different proof using elementary topology only!
 A: Let us continue your approach, assuming by contradiction that $K_n$ is not connected for some $n$. Cover $K_{4n}$ with finitely many open balls of radius $1/(4n)$ and take a connected component $C$ of the union of these balls. Note that this union $B$ is not connected, since a descomposition of $K_n$ yields a decomposition of $K_{4n}$ and of $B$.
Let $E_1$ be the connected component of $\partial C$, which is a closed Jordan curve. Then 
$E_1$ lies entirely in $D$ or entirely in $\bar D^c$. By the Jordan curve theorem, we have a disjoint union $\Bbb{C}=E_1\cup A_1\cup A_2$, where $A_1$ and $A_2$ are open. Each of the open sets $A_1,A_2$ contains points of $\partial D$, so each contains
points of $D$ and of $D^c$. Hence, if $E_1\subset D$, then $D^c$ is not connected, and if 
$E_1\subset \bar D^c$, then $D$ is not connected.
This contradiction shows that $\partial D$ is connected.
If you want to avoid using the Jordan curve theorem, you have to prove a weak version of it, for curves that are part of the boundary of a union of balls of the same radius. 
$\bf{Theorem}$ (weak Jordan curve theorem for boundaries of union of balls): Assume you have a covering of the plane by $\varepsilon$ balls, such that two of them intersect if and only if their centers can be connected by a straight line that intersects no other ball of the covering.  Take $B$ to be the union of a finite number of this balls. Then there is a connected component of $\partial B$, such that we have a disjoint union $\Bbb{R}^2= A_1\cup A_2\cup \partial B$, where $A_1$ and $A_2$ are open, and $B\subset A_1$. 
