A motivating example showing "Every open set can be written as countable Union of disjoint open intervals" I've seen many proofs of this theorem. But, unable to think any example showing this.
Suppose, how to write (0,1) as a countable Union of disjoint open intervals. 
No idea! I'm stuck 
Plz help!
 A: Well, $(0;1)$ is already an open interval. So it is even a union of finitely (namely just one) open interval.
Otherwise, you can also write
$$ (0;1) = (0;1) \cup \bigcup_{n\in \mathbb{N}} \emptyset $$
if you desperately want to have a union over infinitely many things :)
A: Your example $(0,1)$ has only one expression as the union of disjoint open sets; i.e. itself. This is obvious because the union any collection of disjoint open intervals is not connected. 
There is a more general result. Every open set in $\mathbb R^n$ can be expressed in one and only one way as a countable disjoint union of open connected sets.
Sketch: 
$1).\ $ Say $x\sim y$ if there is a connected subspace of $\mathbb R^n$ that contains $x$ and $y$. Check that $\sim$ is an equivalence relation. Then define the components of $\mathbb R^n$ to be the equivalence classes. Take any open set $U\subseteq \mathbb R^n$.The fact that the components of $U$ are maximally connected disjoint sets follows from the definition.
$2).\ $ Components of $U$ are open sets. This follows from the fact that the basis elements in $\mathbb R^n$ are connected. If $x\in C,$ a component of $U,$ then there is a ball $B(x)$ contained in $U.$ And $B(x)\subseteq C$ because $C$ is maximal.
$3).\ $ For uniqueness, note that the $\mathscr A=\{C_{\alpha}\}_{\alpha\in J}$ is an open cover of $\mathbb R^n$. Since $\mathbb R^n$ is Lindelof, $\mathscr A$ has a countable subcover. But the $C_{\alpha}$ are $disjoint$, so the countable subcover must be $\mathscr A$ itself.
