Let $M$ be homeomorphic to a 2-sphere with a finite number $\geq 3$ of points removed. This implies that $M$ can be equipped with a complete, finite area hyperbolic metric. I imagine $M$ as an ideal geodesic polygon in the hyperbolic space with identified edges.
Let $\beta: [0,1] \rightarrow M$ be a closed geodesic in $M$ which we allow to intersect itself.
What I want to show: The complement of the image of $\beta$ in $M$ has only finitely many connected components.
As already answered in this question, it is generally not true that the complement of any closed curve has only finitely many connected components.
Idea: So I was thinking that, maybe the property of $\beta$ being a geodesic implies that geodesic polygons completely bounded by $\beta$ can't have an arbitrarily small area. Then at least it would follow that there can only be finitely many connected components which are completely bounded by $\beta$ (this would actually suffice for my purpose).
Any help is much appreciated :)
Yes, this is true and follows from the following very general
Lemma. Let $\gamma: [0, a]\to M$ be a geodesic in a Riemannian manifold ($M$ need not be complete). Then $\gamma$ has only finitely many self-intersections, i.e. if we define the subset $E\subset [0,a]$ consisting of points $t\in [0,a]$ such that there exists $s\in [0,a]\setminus \{t\}$ such that $\gamma(t)=\gamma(s), \gamma'(t)\ne \gamma'(s)$, then $E$ is finite.
Proof. I will give a proof only in the case when $M$ is a complete hyperbolic surface (which is what you care about) and leave you to prove the general result in case you are interested. Let $X\to M$ denote the universal cover of $M$ with the pull-back hyperbolic metric, which means that $X$ is the hyperbolic plane. Let $G$ denote the group of covering transformations for the universal cover $X\to M$. Then $G$ acts properly discontinuously on $X$. Let $\tilde\gamma: [0,a]\to X$ be a lift of $\gamma$; then $\tilde\gamma$ is again a geodesic. Since $X$ has no conjugate points, the map $\tilde\gamma$ is 1-1. Let $L$ denote the image of $\tilde\gamma$; it is a compact subset of $X$. Therefore, $t\in E$ if and only if there exists $g\in G$ such that $g(L)\cap L$ is nonempty and this intersection is transversal. Transversality (and the fact that $X$ has no conjugate points) implies that the intersection $g(L)\cap L$ consists of exactly one point. Therefore, cardinality of $E$ is at most twice the cardinality of the subset $$ G_L=\{g\in G: gL\cap L\ne\emptyset\}. $$ Since the action of $G$ on $X$ is properly discontinuous and $L$ is compact, the set $G_L$ and, hence, $E$, has finite cardinality. qed
In view of this lemma, the image of $\gamma$ in $M$ is a finite graph $T$ with geodesic edges. Then it is clear that $M\setminus T$ has only finite number of connected components (the closure of each component contains at least one edge of $T$ and closures of only two components can contain the same edge).