Is this argument invalid? So, what I was trying to prove is: Let $\pi : X \to Y$ the quotient map such that $\pi^{-1}(\{y\})$ connected for all $y \in Y$. Then $X$ is connected. 
Suppose yet that $X$ is locally connected. Then $Y$ is locally connected.
Here is how I tried to do this:
For the first part:
$$X = \cup_{y \in Y}\pi^{-1}(\{y\}).$$
Every $\pi^{-1}(\{y\})$ is connected. I don't know if this is right but:
Is true that $\cap_{y\in Y} \pi^{-1}(\{y\}) \neq \emptyset?$
If it is it's done, right?
For the second part, suppose $X$ is locally connected then once $\pi$ is continuous and opened we have $\pi(X) = \cup_{y}{y} = Y$ is locally connected.
 A: HINT: For the first question suppose that $U\subseteq X$ is such that $U$ is both open and closed.


*

*Use the fact that $\pi^{-1}[\{y\}]$ is connected for each $y\in Y$ to show that for each $y\in Y$, either $\pi^{-1}[\{y\}]\subseteq U$, or $\pi^{-1}[\{y\}]\subseteq X\setminus U$.

*Then use the fact that $Y$ is connected and $\pi$ is a quotient map to show that either $\pi^{-1}[\{y\}]\subseteq U$ for every $y\in Y$, or $\pi^{-1}[\{y\}]\subseteq X\setminus U$ for every $y\in Y$. Conclude that either $U=X$ or $U=\varnothing$.

*Explain why this shows that $X$ is connected.
For the second question, I suggest first proving the following result.

Theorem. $X$ is locally connected if and only if each connected component of each open set in $X$ is open.

Then let $U$ be open in $Y$, and show that if $X$ is locally connected, then every component of $U$ is open. Start by letting $C$ be a component of $U$, and let $x$ be any point of $\pi^{-1}[C]$. Let $C_x$ be the component of $x$ in $\pi^{-1}[U]$. The set $\pi^{-1}[U]$ is open in $X$ (why?), so by the theorem $C_x$ is open in $X$.


*

*Show that $\pi[C_x]\subseteq C$.

*Conclude that $$\pi^{-1}[C]=\bigcup_{x\in\pi^{-1}[C]}C_x$$ and hence that $\pi^{-1}[C]$ is open in $X$.

*Explain why this shows that $C$ is open in $Y$.

*Finally, use the theorem again to conclude that $Y$ is locally connected.
