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? 

I think where you're getting confused is that $A_1$ and $A_2$ are open subsets of $S$, but just because something is open in a subspace of $\mathbb{R}$ doesn't mean it's open in $\mathbb{R}$.  Consider the set $E = [0,1)$.  We know that $(-1,.5)$ is open in $\mathbb{R}$ hence 
$$
(-1,.5) \cap [0,1) \;\; =\;\; [0, .5)
$$
is an open subset of $E$, but it is clearly not open in $\mathbb{R}$.  Your proof by contrapositive is correct though, regardless of what $(-\infty, z) \cap S$ looks like in $\mathbb{R}$.  
