If $C$ is a simple closed curve lying in a simply connected open set $U$, then its interior also lies in $U$ 
Let $U$ be a simply connected open set in $\mathbb{R}^2$. If $C$ is a simple closed curve (a space homeomorphic to unit circle $S^1$) lying in $U$, then each bounded component of $\mathbb{R}^2\setminus C$ also lies in $U$.

This is an exercise from Munkres' topology, I don't know how to use the hypothesis that $U$ is simply connected, thanks for any help.
 A: Arguing by contradiction, suppose there exists $x$ contained in the bounded component of $\mathbb{R}^2 \setminus C$ such that $x \not\in U$. 
Choose a base point $p \in C$ and parameterize $C$ by a closed path $\gamma$ based at $p$ and going exactly once around $C$. This closed path represents an element of $\pi_1(U,p)$ which I'll denote $[\gamma]_U$. 
The key fact to use is that $\pi_1(\mathbb{R}^2 - \{x\},p)$ is an infinite cyclic group and $\gamma$ represents a generator of that group. I'll denote that generator $[\gamma]_{\mathbb{R}^2- \{x\}}$.
We also know that the inclusion $i : U \to \mathbb{R}^2$ induces a homomorphism $i_* : \pi_1(U,p) \to \pi_1(\mathbb{R}^2,p)$. 
Since $i \circ \gamma = \gamma$, it follows that 
$$i_*\bigl([\gamma]_U\bigr) = [\gamma]_{\mathbb{R}^2- \{x\}}
$$
Since the right hand side is a nontrivial element of the group $\pi_1(\mathbb{R}^2,p)$, it follows that $[\gamma]_U$ is a nontrivial element of the trivial group $\pi_1(U,p)$, a contradiction.
A: Taking complement, it suffices to show that $U^c$ lies in the unbounded component of $C$. Since $C$ is a simple closed curve, there is a homeomorphism $g:S^1\to C$ from $S^1$ to $C$. Now we use the assumption that $U$ is simply connected to conclude that the map $g:S^1\to U$ is path-homotopic (with range in $U$) to a constant map, hence nulhomotopic. Since $g$ is a homeomorphsim, we see that the inclusion map $i:C\to U$ is also nulhomotopic. Let $x\in U^c$ be a point not in $U$. Note that the map $i:C\to \mathbb{R}^2\setminus\{x\}$ is continuous injective, and nulhomotopic (note that $\mathbb{R}^2\setminus\{x\}$ contains $U$), it follows from Borsuk lemma (see Munkres' topology Lemma 62.2) that $x$ lies in the bounded component $\mathbb{R}^2\setminus C$.
