Proposition: Let $f: X \to Y$ be a quotient map. If the connected components of $X$ are open, then also the connected components of $Y$ are open.
Proof:
Let $y \in Y$ and let $C_y$ be the connected component of $Y$ containing $y$. I want to show that $C_y$ is open. Since $f$ is a quotient map, it will follow from the fact that $f^{-1}(C_y)$ is open.
I want to show that $f^{-1}(C_y) = \bigcup\{C_x \mid x \in A\}$ for some $A \subseteq X$. From this we will have that $f^{-1}(C_y)$ is open being union of open sets.
Let $x \in f^{-1}(C_y)$, then $\exists \, z \in C_y$ s.t. $f(x)=z$ and $C_z=C_y$. Clearly $z=f(x) \in f(C_x)$ and $f(C_x)$ is connected. Then $f(C_x) \subseteq C_z=C_y$, and then $C_x \subseteq f^{-1}(C_y)$.
Doubts
- Is my proof correct?
- I was wondering if $A=f^{-1}(y)$ but I'm not really sure about that...
As said in my comment, your proof shows that you can take $A = f^{-1}(C_y)$.
But it's not generally true that you can take $A = f^{-1}(y)$. For example, let $$X = [-2,-1) \cup [0,+1] $$ and let $$Y = [0,+1] $$ and let $f : X \to Y$ be the quotient map $$f(x) = \begin{cases} x+2 & \quad\text{if $x \in [-2,-1)$} \\ x & \quad\text{if $x \in [0,+1]$} \end{cases} $$ For $y=1 \in Y$, taking $A = f^{-1}(y) = \{1\} \subset X$ will not work.