We will proceed both sides of the implication by proving the contrapositive, i.e., that if the interval property doesn't hold, then the set isn't connected, and conversely, that if the set isn't connected, the interval property doesn't hold.

Suppose $x,y\in E$ and $z\in (x,y),z\notin E.$ Then $E=A_z\cup B_z,$ where

$$A_z=E\cap (-\mathrm{\infty },z),B_z=E\cap (z,\mathrm{\infty }).$$

Since $x\in A_z$ and $y\in B_z,$ they are nonempty, and since $A_z\subset (-\mathrm{\infty },z)$ and $B_z\subset (z,\mathrm{\infty }),$ they are separated. Therefore, $E$ is not connected.

Conversely, suppose, for the sake of contradiction, that $E$ is not connected. Then there are nonempty separated sets $A$ an $B$ such that $A\cup B=E.$ Let $x\in A,y\in B$ and assume Define

$$z=sup(A\cap [x,y]).$$

By Let $E$ be a nonempty set of..., $z\in \bar{A},$ and because $A$ and $B$ are separated, $z\notin B.$ Therefore

If $z\notin A,$ it follows that and $z\notin E.$

If $z\in A,$ then $z\notin \bar{B},$ hence there exists $z_1$ such that and $z_1\notin B$ (because $z\notin \bar{B}$ means there is a neighborhood of $z$ that contains no points of $B$.) Thus, and $z_1\notin E.$