Every open set S in $R^n$ can be expressed in one and only one way as a countable disjoint union of open connected sets. In the following proof in my book their is something I don't understand I will present the proof then present the statement which I don't quite agree with. I want to also ask about some stuff I don't quite understand.
First of all couple of definition:
Component of a of a metric space S is defined as follows:
Suppose x $\in$ S. A component of S is the maximal connected subset of S which contains x.
Theorem 1:
Every point of a metric space S belongs to a uniquely determined
component of S. In other words, the components of S form a collection of disjoint sets whose union is S.
Proof of original theorem:
By theorem 1 the components of S form a collection of disjoint set whose union is S. Each component of T of S is open, because if x $\in T$ then there is an n-bakl B(x) contained in S. Since B(x) is connected, B(x) $\subset$ T, so T is open. By Lindelof theorem the components of S form a countable collection, and by theorem 1 the decomposition into component is unique.
So first of all I don't understand how come B(x) is connected secondly how did they invoke here the lindelof theorem lindelof theorem states for every covering of S there exist a countable covering but how did we know that components are countable covering of S ? 
Lastly I don't understand this theorem quit well I can't picture the statement properly in my head so if someone could explain that it would be great!
 A: For any point $x' \in B(x)$, the line segment from $x'$ to $x$ is entirely contained in $B(x)$. Therefore, $B(x)$ is path connected, and hence connected.
The collection of components of $S$ forms an open cover of $S$, so by Lindelof, it must have a countable subcover. But the components of $S$ form a partition of $S$, so there cannot be any nontrivial subcovers of $S$, i.e. the subcover must be the original cover. (If you remove any component, you'll no longer have a cover of $S$.) Hence, there must be only countably many components.
Finally, for some intuition about the theorem statement, it might help to think about the case $n = 1$. The simplest example of an open subset of $\mathbb{R}$ is of course an open interval $(a, b)$. (In one dimension, an open connected set is just an open interval.) The natural more complicated example is a disjoint union of a sequence of open intervals $(a_1, b_1) \cup (a_2, b_2) \cup \dots$. The theorem makes two claims:


*

*By considering countable disjoint unions of open intervals, we've already exhausted all of the open subsets of $\mathbb{R}$.

*Given only an open set, you can recover the countable set of disjoint intervals from which it was constructed.

