Components are clopen in a space with a finite number of connected components I'm having trouble understanding why this fact is true. A lot of sites just assume it with out reason and it doesn't seem so direct to me. 
Anyways, here is the theorem:

For any topological space $(X,T)$ with finite number of components, each component is clopen.

I only know the most basic definitions of components, basically that it is a maximal connected subset around a point. 
Any help is appreciated. 
 A: A minor variant: since the connected components of $X$ form a partition of $X$ and there is by assumption a finite number of them we can write
$X=C_1\cup \cdots\cup C_n$. Since e.g. $C_1$ is closed, $X\setminus C_1= C_2\cup\cdots\cup C_n$ is open. On the other hand since each $C_i$ is closed the finite union (and this is what goes wrong for an infinite number of connected components) of closed sets $C_2\cup\cdots\cup C_n$ is closed. Hence $C_1$ is open as well. So we proved that $C_1$ is both closed and open, but this clearly works for any of the connected components $C_i$. 
A: We know that component in a topological space $X$ are always closed sets. Also, components form a partition of $X$. So if $X$ has only a finite number of components. Then complement of any component (say) $C_{i}$ in $X$ is still a union of a finite number of remaining components (i.e., the union of a finite number of closed sets) is closed set implies that $C_{i}$ is an open set. Since $C_{i}$ is any component and so each component is clopen set in X.
A: A connected component $C$ of a space $X$ is always closed. This follows from the following fact:

Lemma: If $A$ is a connected subset of a topological space $X$, and $A\subseteq B\subseteq \overline A$, then $B$ is connected as well.

Now if $X$ has only finitely many components, then each component is complement of finitely many closed sets.
A: As @wckrnholm commented is enough.
But I would like to add some more details so that the op can understand better.
The component of a point  $x$ in a topological space is the union of all connected subspace of $X$ which contain the point $x$.
Now the facts

*

*Union of connected subspaces of a topological space is conncetd if they have a point in common.

*If a subspace $S$ of $X$ is connected then every subspace $E$ of $X$ which satisfies  $S\subset E\subset \overline{S}$ also is connected.

imply that components are connected closed subsets of $X$.
Now as @wckrnholm commented

Each component is closed. Hence, the complement of each component is open. But each component is itself a finite intersection of complements of components, hence is also open.

