How bad can a circle domain get? Let $X$ be a domain in the Riemann sphere $\widehat{\mathbb{C}}$. We say that $X$ is a circle domain if every connected component of its boundary is either a circle or a point.
It was conjectured by Koebe in 1909 that circle domains represent all planar domains up to conformal equivalence. This was proved by Koebe himself in the case of finitely many boundary components, and by Zheng-Xu He and Oded Schramm in 1993 in the countable case.
I'm trying to find examples of "bad" circle domains. In particular :
Question : Is there a circle domain whose boundary is not the union of countably many circles, countably many Cantor sets and countably many points?
I'm mostly interested in the case where the boundary has zero area.
Of course, the boundary of a circle domain contains at most countably many circles. The question is whether the set of point components can be written as a countable union of closed totally disconnected sets. The problem seems to be the set of point components that are accumulation points of circles...
Another question :
Question 2 : If $X$ is a circle domain, must there exist an element $z \in \partial X$ which is isolated from circles (in other words there is no sequence of circles in $\partial X$ converging to $z$)?
The answer is yes if the set of point components in closed, by Baire Category considerations...
Thank you,
Malik
 A: Let $K$ be the boundary of your circle domain $\Omega$. 
Let us suppose that every point of $K$ is accumulated on by a sequence of (pairwise different) circle components. Such an example is easy to construct (see e.g. the already existing answer to your Question 2, or simply add the circles inductively - see below for the details).

Claim. A countable union of closed and totally disconnected subsets of $K$ cannot contain all non-circle components of $K$. 

Remark. This shows that the answer to your first question is negative.
Proof. First, suppose that $A$ is a totally disconnected closed subset of $K$, and that $U$ is some open set that intersects one of the circle components, say $C$. Then for any point $z$ on any arc of $C\setminus A$, there are arbitrarily small circle components accumulating on $z$. If they are small enough, then these components are themselves in the complement of $A$, and then there is also a small clopen (in $K$) neighbourhood of each of these components that is contained in the complement of $A$.
So, in summary: $U$ contains a clopen subset $X$ of $K$, which contains a circle component. Furthermore, the diameter of $X$ can be chosen as small as we wish. 
Now if $(A_i)$ is a sequence of totally disconnected closed subsets of $K$, then we can proceed inductively: Find a nonempty clopen subset $X_1$ in the complement of $A_1$ as above, say having diameter less than $1$. Then find a nonempty clopen subset $X_2\subset X_1 \setminus A_1$ of $K$, having diameter less than $1/2$, etc. 
Since each $X_k$ is clopen, and the diameters tend to zero, their (non-empty) intersection is disjoint from all circle components, and belongs to the complement of the union of all $A_i$. This completes the proof.
The proof in fact never used that the non-trivial sets in question are circles. Hence it shows:
Proposition. Let $K\subset\mathbb{C}$ be a compact set such that, for every $\newcommand{\eps}{\varepsilon}\eps>0$, there are only finitely many connected components of $K$ of diameter at least $\eps$. Assume furthermore that every point $z\in K$ is accumulated on by non-point components not containing $z$. Then any countable collection of closed and totally disconnected subsets of $K$ must omit some point components of $K$.
Remark. A slight adaptation of the proof shows furthermore that the set of omitted point components has the cardinality of the continuum.
EDIT. For completeness, let me outline the elementary construction of the domain in question, which provides more details about Misha's "Sierpinski carpet" suggestion. At each stage of the inductive construction, we have a collection $\mathcal{C}_k$ of finitely many pairwise disjoint circles, with $\mathcal{C}_{k+1}\supset \mathcal{C}_k$. 
For each circle $C\in \mathcal{C}_k$, we also pick an open annulus $A_k(C)$ that surrounds $C$, separates $C$ from all other elements of $\mathcal{C}_k$, and whic becomes closer and closer to $C$ as $k\to\infty$. 
Start with $\mathcal{C}_0$ having one circle $C$ in it, with some annulus $A(C)$ picked around it. Then, inductively, for every circle $C$ in $\mathcal{C}_k$, add a whole bunch of small circles to $\mathcal{C}_{k+1}$, within the inner curve of $A_k(C)$, such that every point of $C$ is close to one of these new circles. Then pick the annuli with the desired property, and so that they are pairwise disjoint. 
Let $K$ be the closure of the union of all these circles. The construction ensures that this set is totally disconnected (any two points will be separated by one of the annuli), and clearly $K$ has all the desired properties).
A: For the 2nd part one can construct an example by imitating the Sierpinski Carpet construction. Or, you use Kleinian groups. Take two Fuchsian subgroups $F_1, F_2$ of $PSL(2,C)$ with disjoint limit circles. After replacing them with finite index subgroups if necessary, the subgroup $G$ of $PSL(2,C)$ that $F_1, F_2$ generate is a functional group, i.e. its domain of discontinuity contains a connected $G$-invariant component $\Omega$; $\Omega$ is easily seen to be a circular domain. At the same time, due to minimality of the action of $G$ on its limit set (which is the boundary of $\Omega$), each point of the limit set is the limit point of a sequence of circles whose radii tend to zero. You can find more details on such constructions (called "Klein combination") in Maskit's book "Kleinian Groups". He also gives a detailed background on Kleinian groups. 
Edit: I have to think more about Part 1, I am not longer sure if it also has negative  answer.     
A: Ok, after writing many comments thinking I had a proof showing that Question 1 had a negative answer, I now believe that the following construction (inspired by Misha's answer) yields an example. I hope I am not mixing everything and that what I write makes sense.
Take the usual Cantor ternary set $K$ in $[0,1]$ (built by removing middle intervals) and embed it in the horizontal axis in $\mathbb{C}$. Say that $x\in K$ is an endpoint if it is the endpoint of an interval in some stage of the construction of $K$. Replace these endpoints by small circles that do not intersect or encircle any part of the rest of $K$. You can do that because there is some room on one side of any endpoint, which my be thought as the westmost or eastmost point in the added circle. The circles radii go to zero when you go "further down" in $K$, so you do not add new accumulation points, and the resulting subset $L\subset\mathbb{C}$ is compact. Then $L$ is the boundary of the domain $\mathbb{C}-L-\{\text{interior of the circles}\}$. If you remove the circle boundaries, you obtain the Cantor set minus its endpoints, which is homeomorphic to the irrationals (or the Baire space $\omega^\omega$, see for instance the answers to this question: https://math.stackexchange.com/questions/52073/two-questions-about-the-cantor-set-construction), which cannot be a countable union of Cantor sets.
