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).
