Connected Components of Locally Path-Connected Spaces are Open
# Statement
Let $X$ be a Topological Space that is Locally Path-Connected. Then every Connected Component $C \subset X$ is Open.
# Proof
Let $C \subset X$ be a Connected Component of $X$. Since $X$ is Locally Path-Connected, for each $x \in C$, there exists $U \subset X$ Open so $x \in U$ and $U$ is Path-Connected. Recall Path-Connected implies Connected, so $U$ is Connected. Since $U \cap C \ni x$, we know that $U \cup C$ is Connected because If Connected Sets share a point, then their Union is Connected. But $U \cup C \supset C$. Because $C$ is Maximal, $U \cup C = C$. Thus $U \subset C$. Since $x$ was arbitrary, we can find such a $U_{x}$ for each $x \in C$. But then $$C = \bigcup\limits_{x \in C}U_{x}$$so $C$ is Open. $\blacksquare$