A connected component of a locally Euclidean space is open

Question

I want to prove that a connected component of a locally Euclidean space X is open in this space.

I start the proof taking a point y in the connected component Y of X. In particular, y is a element of X and have an open neighborhood U, and there is an open subset in an euclidean space and a homeomorphism. I'm thinking on taking the intersection of this neighborhood with Y, defining an open ball contained in the image of this intersection under the homeomorphism and taking it back with the inverse function, but is not clear for me how to use the fact that Y is connected.

Answer 1

Let $C$ be a component. By local connectedness, for
all $x$ in $C$, some open connected $U(x)$ with $x \in U(x)$.
Show for all $x$, $U(x) \subseteq C$ and $C = \cup\{ U(x) : x \in C \}$.

Answer 2

If $X$ is (weakly) locally connected (in the sense that every point has a connected neighbourhood) then every component of $X$ is open.

Proof: Let $C$ be a component of $X$, i.e. a maximal connected subset of $X$. Let $x \in C$. Then let $U_x$ be a connected neighbourhood of $x$. Then $C \cup U_x$ is connected as the union of two connected intersecting subsets (as $x \in U_x \cap C$). As $C \subseteq C \cup U_x$ ,by maximality of $C$ we have that $C = C \cup U_x$ which implies $x \in U_x \subseteq C$. This shows that $x$ is an interior point of $C$ and as $x \in C$ was arbitrary, $C$ is open.

Now, let $X$ be locally Euclidean. So for every $x \in X$, we have a neighbourhood $U_x$ of $x$ and a homeomorphism $h: U_x \to U$ where $U$ is an open subset of $\mathbb{R}^n$. Now $y=h(x) \in U$ has a standard Euclidean ball neighbourhood $B(y,r) \subseteq U$ for some $r>0$, and it's a standard fact that $B(y,r)$, as all open balls in $\mathbb{R}^n$, is connected. But then $h^{-1}[B(y,r)]$ is a connected neighbourhood of $x$ in $X$, and so the previous fact applies, and all components of $X$ are open.