If a closed set does not separate $\mathbb{R}^n$, can one of its connected components separate $\mathbb{R}^n$?

More generally, suppose $X$ is any connected space and $F\subseteq X$ is a closed subset such that $X\setminus F$ is connected. Then I claim that for any connected component $C$ of $F$, $X\setminus C$ is connected. To prove this, suppose $X\setminus C$ were disconnected, so it is a disjoint union of two nonempty open subsets $U$ and $V$ (which are also open in $X$ since $C$ is closed in $X$). Since $X\setminus F$ is connected, it is completely contained in one of $U$ and $V$; let us say $X\setminus F\subseteq U$. Then $V\subseteq F$ and thus $C\cup V$ is disconnected since $C$ is a connected component of $F$. So, we can write $C\cup V$ as a disjoint union of two nonempty closed subsets $A$ and $B$. Since $C$ is connected, it is contained in one of $A$ and $B$; say $C\subseteq A$, so $B\subseteq V$. Then I claim $B$ is clopen in $X$, contradicting the assumption that $X$ was connected. First, $A$ and $B$ are both closed in $X$ since they are closed in $C\cup V$ and $C\cup V=X\setminus U$ is closed in $X$. Then $B$ is also open in $X$ since its complement is $(X\setminus V)\cup A$, a union of two closed subsets of $X$.