I am reading Colin Buxton's proof of the no retraction theorem in 2 dimensions on pages 9 and 10 here: http://math.stmarys-ca.edu/wp-content/uploads/2017/07/Colin-Buxton.pdf.
There is one step I am not convinced of. First I introduce the notation relevant to my question.
$\textbf{NOTATION:}$
$\mathbb{D} = \{x \in \mathbb{R}^2 : ||x||\leq 1\}$,
$C=\{x \in \mathbb{R}^2 : ||x||=1\}$,
$r:\mathbb{D} \rightarrow C$ is a retraction (which we assume exists for contradiction),
$a, b \in C$ with $a \neq b$,
and $A=r^{-1}(\{a\}), B=r^{-1}(\{b\})$.
$\textbf{MY CONFUSION:}$
Buxton says: "Note that $\overline{C\diagdown \{a, b\}}=C$. We can, then, find a subset of $\mathbb{D}\diagdown (A \cup B)$ whose closure will contain $C$. Let us call this set $P$. We can choose $P$ to be open and $\textit{path connected}$."
$\textbf{How can we choose $P$ to be path connected?}$
It is clear to me that we can choose $P$ to be open and contain $C$ in its closure. For example, let $B = \{x \in \mathbb{R}^2 : ||x||<1\}$. Then clearly $P=B \diagdown (A \cup B)$ will be open and $C \subseteq \overline{P}$. However, no reason jumps out at me why $P$ can be chosen to be path connected.
I am probably missing something terribly simple.
Thank you!