Let $\alpha$ be a path in the Euclidean plane $\mathbb{R}^2$. Let $A, B$ be distinct connected components of $\mathbb{R}^2\setminus\operatorname{Im}\alpha$. Let $\beta$ be a path in $B$. Let $C$ be a connected component of $\mathbb{R}^2\setminus\operatorname{Im}\beta$. Is it necessarily the case that either $A\cap C=\emptyset$ or $A\subseteq C$?
P.S. Even though I tagged this question with "algebraic-topology", I don't know algebraic topology, and therefore would appreciate as elementary a reply as possible. You can use the Jordan Curve Theorem, and the fact that every open set of $\mathbb{R}^2$ is the disjoint union of a countable number of open, connected sets.
Answer Yes, it is necessarily the case that either $A\cap C=\emptyset$ or $A\subseteq C$.
I will prove a more general statement (cf. the Remark below):
Restatement of the problem Let $T = \langle\Omega, \tau\rangle$ be a locally-connected space. Let $A \subseteq \Omega$ be connected, and let $F$ be a closed subset of $\Omega\setminus A$. Let $C$ be a connected component of $\Omega\setminus F$. Then $A\cap C=\emptyset$ or $A\subseteq C$.
Proof Since $\Omega\setminus F$ is open, $\Omega\setminus F$ is locally connected. Denote by $\mathbf{C}$ the set consisting of the connected components of $\Omega\setminus F$. Then $\mathbf{C}$'s members are open and pairwise disjoint. Define $D = \cup\big(\mathbf{C}\setminus\{C\}\big)$. Then $C$ and $D$ are open and mutually disjoint, and $\Omega\setminus F = C\cup D$. Since by assumption $F$ and $A$ are mutually disjoint, $A \subseteq \Omega\setminus F$. Hence, $A \subseteq C\cup D$. Then, since $A$ is by assumption connected, either $A \subseteq C$, or $A \subseteq D$. $\square$
Remark
If $T$ is moreover Hausdorff, then $F$ can be taken to be the image of an arbitrary path $\beta$ in $\Omega\setminus A$. In fact, since $\beta$ is a path, it is by definition a continuous function of the compact set $[a,b] \subseteq \mathbb{R}$ (w.r.t. the Euclidean topology on $\mathbb{R}$) for some $a, b \in \mathbb{R}$ such that $a <b$. Therefore its image $\operatorname{Im}\beta$ is compact. Since $T$ is Hausdorff, $\operatorname{Im}\beta$ is closed.
Since the Euclidean space $\mathbb{R}^2$ is Hausdorff, as well as locally connected, the restated proposition resolves the original question in its original formulation.