Can the complement of a simply connected set in $\bar{\mathbb{C}}$ in an open set always be covered by a simply connected union of balls? I believe the following to be true, but am worried my intuition does not account for fractally things:
Let $K\subset\bar{\mathbb{C}}$ ($\bar{\mathbb{C}}$ being the Riemann sphere) be closed (thus compact) and suppose that $K$ and $K^c$ are both connected (thus $K$ is the complement of an open simply connected sub-set of $\bar{\mathbb{C}}$), and let $U\subset\mathbb{R}^2$ be any open set containing $K$.  Then there is a list of balls $B_1,\ldots,B_N\subset U$ such that $\displaystyle K\subset\bigcup_{i=1}^NB_i$ and $\displaystyle\bigcup_{i=1}^NB_i$ is simply connected.
I thought this would be straight forward but am not so sure now.
 A: I figured it out; we do it in two steps.
Assume that $K$ is bounded.  First cover $K$ in finitely many balls $B_1,B_2,\ldots,B_N$, all contained in $U$ (easy application of compactness and openness).  It is easy to see that a union of finitely many balls has at most finitely many bounded components of its complement. Let $D$ be some bounded component of
$$\left(\displaystyle\bigcup_{i=1}^NB_i\right)^c.$$
Draw a path from some point in $D$ to $\infty$ which does not intersect $K$.  In the Riemann sphere, this path is compact, so we may cover this path with finitely many balls $C_1,\ldots,C_M$.  Define
$$K'=\left(\displaystyle\bigcup_{i=1}^NB_i\right)\setminus\left(\bigcup_{i=1}^MC_i\right).$$
Do this for each of the finitely many bounded components of $$\left(\displaystyle\bigcup_{i=1}^NB_i\right)^c.$$
Then we obtain a new set $K'\subset U$ which is the union of finitely many balls minus finitely many balls which has the same properties as $K$, namely that $K'$ is compact and both $K'$ and ${K'}^c$ are connected.  Note also that the boundary of $K'$ consists of finitely many smooth pieces.
Now that $\partial K'$ is piecewise smooth, with finitely many "pieces", it is easy to show that this $K'$ has the desired property, and whichever cover works for $K'$ will also work for $K$.
