Theorem 1   A topological space $X$ is locally connected if and only if each component of an open set is open.

Proof. First, suppose that $X$ is locally connected and that $U$ is an open set of $X$ Let $p \in C$ where $C$ is a component of $U$ Since $X$ is locally connected there is an open connected set, say $V$ with $p \in V \subset U$ Since $C$ is a component of $U$ it must be that $V \subset C$ Hence, $C$ is open. For the converse, suppose that each component of each open set is open. Let $p \in X$ Let $U$ be an open set containing $p$ Let $C$ be the component of $U$ which contains $p$ Then $C$ is open and connected, so $X$ is locally connected.