Locally path-connected implies that the components are open If $X$ is a locally path-connected space, then its connected components are open. 
I am trying to prove this, but for some reason it doesn't seem right to me, knowing that components are always closed. If the statement is true, wouldn't it be the case the components are the whole space $X$?

Consider a connected component $C$ of $X$.
(*) Given any point $x \in C$ there must exist an open set $V \subset X$ that is path-connected such that $x \in V$

We say X is locally path connected if for every point $x \in X$ and every open neighborhood $U \subset X$ of $x$, there exists a open set $V \subset X$ that is path-connected and $x \in V \subset U$.

To get (*) just consider $U = X$ in the definition above.
Now, Since $V$ is path-connected then it must bet connected. And since $V \cap C \neq \emptyset$ ($x \in V\cap C$) this means $V \subset C$.
$C$ is open.
