QUESTION: Prove that each connected component of a surface is also a surface. (Hint: Connected components of a locally connected space are open.)
Definition: A space $X$ is said to be locally connected at $x$ if for every neighborhood U of x, there is a connected neighborhood $V$ of $x$ contained in $U$.
Theorem 1: A space $X$ is locally connected if and only if for every open set $U$ of $X$, each component of $U$ is open in $X$.
Theorem 2: Every open subset in a surface is also a surface.
MY ATTEMPT: Let $x\in S$ be a connected component in the surface $S$, such that for every neighborhood $U$ of $x$ there is a connected neighborhood $V$ of $x$ such that $x\in U$, i.e. $S$ is locally connected at $x$. Now, due $U$ is an open subset in $S$, let $C$ be a component of $U$, if $x\in C$ we can choose a connected neighborhood $v$ of $x$ such that $V\subset U$. Because $V$ is connected, it works for every $C\subset U$. Therefore $C$ is open in $S$.
MY DOUBT: Unfortunatelly my attempt is very similar to the proof of theorem 1 (in James Munkres). I just modified few things. My doubts are how can I garantee that always exists a connected neighborhood $V$ of $x$ (because I'm in $S$ wich is a surface) and if my proof provided is correct?