It's been a while since I've touched topology so please bear with me.
Theorem: Every non-empty open set on the real line is the union of countable disjoint class of open intervals.
proof:
Let $G$ be a non – empty open set on the real line. Since $G$ is non-empty, it's order is at least one. Since $G$ is an open set, this implies that for every element $x$ in $G$, there exists a positive radius $r$ such that an open interval $I$ may be centred on $x$ such that $I$ is contained in $G$.
But for each of these open intervals $I_{i}$, there exists some radius $\bar{r}$ such that an open interval $\bar{I_{i}}$ is contained in $I_{i}$. Hence, $I_{i}$ is class of open set.
The union of $\bar{I_{i}}$ then is an open set.
Hence, $I$ is the union of open sets.
If y is another point then $I_{x} = I_{y}$.
Why is this so?
Best Answer
Your solution is a bit unclear, I'm not sure what $I_x, I_y$ are so here's another approach:
Let $G\subseteq\mathbb{R}$ be a nonempty open subset of $\mathbb{R}$. Let $x\in G$. Denote by $C(x)$ the connected component of $x$ in $G$, i.e. the maximal connected subset of $G$ containing $x$.
Proof. Let $a,b\in C(x)$ are such that $a<b$. Assume that $a<c<b$. If $c\not\in C(x)$ then $(-\infty, c)$ and $(c,+\infty)$ would be a nontrivial decomposition of $C(x)$ into open subsets. That contradicts $C(x)$ being connected. Therefore $c\in C(x)$ and so $C(x)$ is an interval. We will show it is open:
Since $G$ is open then there is an open interval $I$ around $x$ such that $I\subseteq G$. Since $I$ is connected and $x\in I$ then $I\subseteq C(x)$ which completes the proof. $\Box$
Proof. Obviously $G=\bigcup_{x\in G} C(x)$ and $C(x)$ are either disjoint or equal. So the only question is: can we take only countably many of them? The answer is yes. That's because for a given $C(x)$ there exists $q\in\mathbb{Q}$ such that $q\in C(x)$ and $C(x)=C(q)$. Indeed, you take an open interval $I$ around $x$ fully contained in $C(x)$ and then you pick any rational from $I$ (rationals are dense). Therefore
$$G=\bigcup_{x\in G\cap\mathbb{Q}} C(x)$$
which completes the proof. $\Box$