Is every planar bounded $C^2$ domain finitely connected? Let $\Omega \subset \mathbb R^2$ be a bounded $C^2$ domain. Is $\Omega$ then finitely connected? As I learned recently a domain in $\mathbb R^2$ is finitely connected iff “[its] complement has finitely many components (holes)“.
The first (mostly smooth) bounded domain with infinitely many holes that comes to mind is
$$\Omega = B_1(0) \setminus \overline{\bigcup_{n=2}^\infty B_{1/n^2}(1/n)},$$
however this domain fails to be $C^2$ at $0$. (Here $B_{r}(x)$ is the ball with radius $r$ centered at $x$.)
Similar constructions should run into the same issue. In particular, suppose that $\Omega$ is not finitely connected and enumerate the holes by $A_1, A_2, \dots$. For $n \in \mathbb N$, take $x_n \in \partial A_n$. As $\partial \Omega$ is compact, there exists a converging subsequence. I doubt that $\Omega$ can be $C^2$ at this limit  point. However, I fail to make this precise.
My main interest is to apply a result holding for finitely connected $C^2$  domains to a bounded domain and thus I would welcome a reference stating that bounded $C^2$ domains are always finitely connected. This is also the reason I require the domains to be $C^2$ but the same question can of course also be asked for Lipschitz domains, for instance.
Edit: A domain is $C^2$ if it can locally be written as the set of points above the graph of a $C^2$ function. A domain is bounded if it is contained in $B_R(0)$ for some $R > 0$. A bounded $C^2$ domain is a domain which is both bounded and $C^2$.
 A: Yes. It is finitely connected. The boundary of your domain is a compact one-dimensional manifold and therefore it consists of a finite number of curves diffeomorphic to circles. You can find a proof of classification of compact one dimensional manifolds in:
J.W. Milnor, Topology from the differentiable viewpoint. Based on notes by David W. Weaver University Press of Virginia, Charlottesville, Va. 1965.
A: I'm not sure what you mean precisely by bounded $C^2$ domain, but I'll take it to mean a compact set with positive reach (that includes slightly weaker regularity domains in fact).
Such domains are indeed finitely many connected. One way to see it is that, for $r$ less than the reach, the $r$-neighborhood of the domain deformation retracts to the domain, hence the inclusion of the domain into its $r$-neighborhood is an isomorphism at the homology level.
Since one can construct a polygonal domain sandwiched between the domain and its $r$-neighborhood, the inclusion map isomorphism in homology described above factors through a finite dimensional space, hence its rank, which is equal to the Betti numbers of the domain, is finite.
