Connected components in a surface

A connected component of a topological space is a closed set because the closure of a connected set is connected again. Consider now $S_i$ a connected component of a surface $S$. Then we prove that $S$ is an open set, in particular, $S_i$ is a surface. This result holds because any surface is locally connected. In order to avoid the use of this terminology, take $p\in S_i$ and we prove that $p$ is an interior point of $S_i$. Consider $X:U\rightarrow V\subset S$ a parametrization around $p$, where $V\subset S$ is an open set around $p$. Since $U$ is an open set of ${\mathbb R}^2$, there exists a ball $B$ centered at $q=X^{-1}(p)$. In particular $B$ is connected and consequently, $X(B)$ is a connected open set around $p$. Since $S_i$ is the connected component of $p$, then $X(B)\subset S_i$, q.e.d.

A consequence: any component of a compact surface is a compact surface. Indeed, if $S_i$ is the component, then $S_i$ is closed and clearly bounded, so $S_i$ is a compact. And by the above paragraph, $S_i$ is an open set of $S$, so it is a surface.