Consider the set $S=\{ (1),(12),(13),(23),(123),(132)\}$. Let $G$ be a group of permutations of a set $S$. Prove that the orbits of the members of $S$ constitute a partition of $S$.
Now
\begin{eqnarray*} \operatorname{orb}(1)&=&\{ 2,3\}\\ \operatorname{orb}(2)&=&\{ 1,3\}\\ \operatorname{orb}(3)&=&\{ 1,2\}\\ \end{eqnarray*}
These sets are not disjoint. Hence my doubt.
Then how to proceed to prove second statement.
Any hint or suggestion will be appreciated.
On
In your example you always forgot that $i \in \text{orb}(i)$ since $(1)(i)=i$ for $i \in {1,2,3}$. Thus all the orbits coincide, and you do not have a counterexample.
For the proof of fact 2): consider two orbits not disjoints, so $a \in \text{orb}(x) \cap \text{orb}(y)$. By definition we have $g,h \in G$ such that $gx=a=hy$, from this you obtain $y=h^{-1}gx$ and now you can easily deduce that the two orbits coincide.
On
An alternative is to consider the map $orb: S \to \mathcal P(S)$ given by $x \mapsto orb(x)$.
Then the relation $x \sim y$ iff $orb(x)=orb(y)$ is clearly an equivalence relation on $S$, because equality is an equivalence relation.
Therefore, the equivalence classes of this relation form a partition of $S$. These equivalence classes are exactly the orbits. This is the only point that needs a little work.
Let $e\in G$ denote the identity element and let $x\in S$. Then by definition of an action you have $e\cdot x=x$, so $x\in\operatorname{orb}(x)$. This is where your supposed counterexample fails. Intuitively speaking; every element can be reached from itself by the action of something in $G$; by doing nothing.
A key step in proving that orbits are disjoint is proving that $$x\in\operatorname{orb}(y)\quad\Leftrightarrow\quad y\in\operatorname{orb}(x).$$ From this it is not hard to show that the relation $\sim$ defined by $$x\sim y\quad\Leftrightarrow\ x\in\operatorname{orb}(y),$$ is an equivalence relation. The equivalence classes partition $S$, and these are precisely the orbits.