I have a doubt with my proof of the following result:
A topological space $(X,\mathcal{F})$ is locally connected if and only if for every open set $U \subset X$ each connected component of $U$ is open in $(X,\mathcal{T})$.
I have tried:
If $(X,\mathcal{F})$ is locally connected, given a component $C$ of an open $U$, for $x \in C$ we can take a connected neighborhood $V \subset U$. Since $V$ is connected it implies $V \subset C$, so $C = \cup_{x \in X}V_x$ is open in $X$.
For the another implication, given a neighborhood $U$ of $x \in X$, suppose $C$ is the component of $U$ containing $x$. By hyptothesis, $C$ is open and connected in $X$, then $(X,\mathcal{F})$ is locally connected in $x$. Since $x$ is generic, is $(X,\mathcal{F})$ locally connected.
Is it right?
Thank you very much.