$E \subset \mathbb R$ is an Interval $\iff E$ Is connected My text gives the definition that $E$ is disconnected if there exist disjoint open sets $A, B$ such that:


*

*$A \cap E$, $B \cap E$ are nonempty.

*$(A \cap E) \cup (B \cap E) = E$.


Then for the $(\implies)$ direction, the text offers the following as proof:
Suppose $E$ is an interval and $E$ is not connected.  Then there exist disjoint open sets $A$, and $B$ with $A \cap E \ne \emptyset$, $B \cap E \ne \emptyset$, and $E \subset A \cup B$.  Suppose $a \in A \cap E$ and $b \in B \cap E$.  Inasmuch as $E$, an open set in $\mathbb R$ is a finite or countable union of open and disjoint intervals, there exist disjoint open intervals $I$ and $J$ such that $a \in I$ and $b \in J$.  Suppose $a < b$ and $J = (t, s)$.  Since $a < t < b$, $t \in E$.  However, $t \notin A \cup B$, creating a contradiction.
Is the part with $E$ being a finite or countable union of open and disjoint intervals necessary?  I did it in a much simpler way:  Without loss of generality, call the open set on the left $A$, with $a \in A \cap E$, and call the open set on the right $B$, with $b \in B \cap E$.  Now, set $c =$ sup$A$;  $a < c < b \rightarrow c \in E$.  Since $c$ is a boundary point of $A$ and $A^c$, $c \notin A$.  But $c \notin B$ either.  Otherwise, there is a $\delta$ such that $N(c, \delta) \subset B$, entailing that $A \cap B$ is nonempty, which is not possible.  As a result, we have $c \in E$ and $c \notin E$, a contradiction.
Why did the book need to use the fact that $E$ is a union of open, disjoint intervals?
 A: Your proposed proof isn't quite right. The problem is that it's not guaranteed that there is a "left" or "right" open set - this would be true if $A$ and $B$ were intervals, but they need not be. The open sets $A$ and $B$ could interlace.
So it's not actually guaranteed that $\sup A \in E$ or that $\sup B \in E$, precisely because of this issue: If we take smaller and smaller intervals in $A$ growing to the right, and likewise for $B$, then it's certainly possible to have
$$\sup A = \sup B = \sup E \notin E$$
Notice that this is fixed precisely by knowing that an open set can be decomposed into intervals; this actually does allow you to select a "left" interval and a "right" interval, and then study how they don't intersect.
A: The book needs the Hausdorff separation property -- that around each of two points there are disjoint open sets.  This property is normally covered in Topology, but your book is importing the idea via $E$ being a finite or countable union of open and disjoint intervals so it can pick out two disjoint intervals as needed. 
