Locally connected

Question

$\textbf{Theorem:}$ $X$ is locally connected if and only if the components of each open subset of $X$ are open.

I have a question in the next proof.

$\textbf{Proof:}$ Suppose that $X$ is locally connected. Let $U$ be an open subset of $X$, and $C$ be a component of $U$. If $x \in C$, then there is a connected open set $V \subseteq X$ such that $x \in V \subseteq U$, by our hypothesis. Since $C$ is a component of $x$ in $U$ and $V$ is a connected subset of $U$ containing $x$, we have $V \subseteq C$. Thus $C$ is a neighbourhood of each of its points, and therefore open.

My question is in the part where they state that $V \subseteq C$.

I know that if we have a topological space $X$ and $Y \subseteq X$ connected (in $X$), then $Y \subseteq C$ where $C$ is a component of $X$.

Now, in the proof we have $V \subseteq U$ and $V$ connected in $X$. We would have to prove that $V$ is connected in $U$ and then we have $V \subseteq C$ because $x \in C$.

In the proof they do not mention any of this, they simply say that $V$ is a connected (in $X$) subset of $U$. This condition implies that $V$ is connected in $U$?

the connectedness of subsets of a subspace confuse me.

Answer 1

$U$ is open, so open sets in $U$ in the subspace topology are precisely those open sets in $X$ which are contained in $U$. It follows that for a subspace of $U$, being connected in $U$ vs. in $X$ are equivalent.

Answer 2

If $A \subseteq X$ is a subspace there is no notion of $A$ being "connected in $X$". Just that $A$ is connected as a space (i.e. it has no non-trivial clopen set). Of course, $A$ does have the subspace topology wrt $X$, but for connectedness (and compactness and local connectedness etc.) we should just judge $A$ as a space on its own.

If $U$ is open, $C$ is a component of $U$ (as its own space), $x \in C$: As $X$ is locally connected, we know there is a connected neighbourhood $V_x$ of $x$ that is a subset of $U$ (this uses two things: $U$ is open in $X$ and the connected neighbourhoods form a local base at $x$ in $X$).

So $V_x$ is connected, it is a subset of $U$ and contains $x$. So looking inside $U$ as its own space, it's by definition that $C$, the component of $U$, obeys $V_x \subseteq C$ by maximality of the component. This shows that any $x\in C$ is an interior point (in $U$, just as in $X$) of $C$, so $C$ is open.