$\mathbb{R}^2 \setminus C$ subspace has two connected components where $C$ is a circle. Let's look at $\mathbb{R}^2$ with usual metric topology and in it circle. Then author says that $\mathbb{R}^2 \setminus C$ subspace has two connected components.
First of all I need to check they are connected components. They are not subset of each other so we need to check if they are connected subsets. In order to prove they are connected subsets we must show that they can't be written union of two non-empty disjoint open sets. Here is where I stuck. I don't know how to show they can't be written like that.
EDIT.
I read and can understand first part of proof of @MoisheKohan , but I am stuck on second part.We know that $Lp∪Lq$ is connected how conclude that A is connected as well?
 A: Since this problem is likely used as a homework, I will only leave a sequence of hints:

*

*First of all, you have to figure out what "they" are in the sentence "First of all I need to check they are connected components." Let's call "them"  simply $A$ and $B$. Once you identify "them," you will have no trouble checking that ${\mathbb R}^2 \setminus C$ is the disjoint union of $A$ and $B$, and that both $A$ and $B$ are nonempty.


*Next, you will prove that one of these subsets, namely $B$, is convex, which means that with every pair of points $p, q\in B$, the line segment $pq$ is also contained in $B$. Now, use the fact that you already know, namely that the (straight) line segments $pq$ are connected, to show that $B$ cannot be written as a disjoint union of two nonempty open subsets. (You will be arguing by contradiction.)


*Then, you proceed to analyze $A$. This subset is not convex. But you will prove that it has the following property:
For every pair of points $p, q\in A$, there are line segments $L_p, L_q$ both contained in $A$ and containing $p, q$ respectively, such that $L_p\cap L_q\ne \emptyset$.
The proof of existence of such line segments can be done in many ways, for instance, using axiomatic Euclidean geometry, or trigonometry, or analytic geometry, or linear algebra.
Afterwards, you argue that $L_p\cup L_q$ is connected. (In a comment you indicated that you already know how to prove that if $Y, Z$ are connected subsets of a topological space and $Y\cap Z$ is nonempty, then $Y\cup Z$ is connected.) Lastly, arguing as in Part 2, you conclude that $A$ is connected as well.
This will conclude the proof.
A: Let us choose the circle $C_r =\{(x,y):x^2 +y^2=r^2\}$
Claim: $\Bbb{R^2}\setminus C_r$ has two connected components , one is bounded and other is unbounded.
$ B(0, r) =\{(x,y):x^2 +y^2<r^2\}$
\begin{align}Ext[B(0, r)]&=\{(x,y):x^2 +y^2>r^2\}\end{align}
$\Bbb{R^2}\setminus C_r=B(0, 1) \sqcup Ext[B(0, r) ]$

Connected component: $(X, \tau) $ be a topological space.Let, $p\in
 X$, then connected component of $X$ containing $p$ is the maximal
connected subsets of $X$ containing $p$.

Let, $p\in \Bbb{R^2}\setminus C_r$ , then either $p \in B(0, r) $ or $p\in Ext[B(0, r) ]$
Case I: $p \in B(0, r) $
Clearly, $ B(0, r) $ is a connected subset of $\Bbb{R^2}\setminus C_r$ containing $p$.
And it is again easy to see it is the maximal connected subset of $\Bbb{R^2}\setminus C_r$ containing $p$.
If $\exists M\subset \Bbb{R^2}\setminus C_r$ properly containing $B(0, r) $, then $M$ contain points $q_i$  of $Ext[B(0, r) ]$.
As $Ext[B(0, r) ]$ is open subsets of $\Bbb{R^2}\setminus C_r$, there exists an open set $U_i$ containing $q_i$.
Then , $U=\cup_{i}U_i\subset \Bbb{R^2}\setminus C_r$ is open .
and
$M=B(0, r) \sqcup U\implies M$ is not connected.
Hence, $B(0, r) $ is a connected component of $\Bbb{R^2}\setminus C_r$ for any point inside it.
Case II: Using similar argument you can prove easily, for any $p\in Ext[B(0, r) ]$ maximal connected subset containing $p$ is the $Ext[B(0, r) ]$.
Combining above two cases, we can say $\Bbb{R^2}\setminus C_r$ has two connected components $B(0, r) $ and $Ext[B(0, r) ]$.
