Proving Any connected subset of R is an Interval Common Proof:

Suppose $S$ is not an interval of $R$.
Then by Interval Defined by Betweenness, $∃x,y∈S$ and $z\in R∖S$ such that $x<z<y$.
Consider the sets $A_1=S∩(−∞,z)$ and $A_2=S∩(z,+∞)$.
Then $A_1,A_2$ are open by definition of the subspace topology on S.
Neither is empty because they contain x and y respectively.
They are disjoint, and their union is S, since z∉S.
Therefore $A_1∣A_2$ is a separation of S.
It follows by definition that S is disconnected.

But why are $A_1,A_2$ open sets? 

Do you know what the subspace topology is?  The larger set is $\Bbb R$ with the usual topology.  Since $S \subseteq \Bbb R$, we can equip $S$ with the subspace topology, i.e., the topology where each $V$ that is open in S is of the form $V = S \cap U$ with $U \subseteq \Bbb R$ open in $\Bbb R$.
Since $A_{1} = S \cap (-\infty, z)$, and the interval $(-\infty, z)$ is open in $\Bbb R$, then $A_{1}$ is open in $S$ with the subspace topology.  Similarly, $A_{2}$ is open in $S$ with the subspace topology.
Please note the following very important fact: we determine whether a set is connected or disconnected based on the open sets in its topology.  What I mean is, if $(X, \mathcal{T})$ is a topological space, and $Y \subseteq X$ is a subset, we determine whether $Y$ is connected or not using the sets in the subspace topology of $Y$.
So, the set $[0,1) \cup (2,3]$ is disconnected.  Why?  Let $A = [0,1)$ and $B = (2,3]$.  Neither $A$ nor $B$ are open in $\Bbb R$, but both of them are open in $[0,1) \cup (2,3]$ under the subspace topology of this set.  Since they are open in the subspace topology, and clearly both are disjoint and nonempty, and $A \cup B$ is the entire set, then $[0,1) \cup (2,3]$ is disconnected. 
