Connected subsets of the real line are intervals

Statement

Let $S \subseteq \mathbb{R}$, where $\mathbb{R}$ has the usual topology. Then $S$ is connected if and only if $S$ is an interval, that is, for all $a, b \in S$ and all $c \in \mathbb{R}$ with $a < c < b$, $c \in S$.

Remarks

If $S$ is not an interval, take $a < c < b$ with $a, b \in S$ and $c \notin S$. Then $(-\infty, c)$ and $(c, \infty)$ split $S$. The proof that an interval $I$ is connected uses the completeness axiom. If open sets $U$ and $V$ split $I$, with $a \in U$, $b \in V$ and $a < b$, let $c$ be the supremum of the points of $U \cap [a, b]$. Then $c$ lies in neither $U$ nor $V$. Under this definition the empty set and single points are intervals, and they are connected. The statement fails for $\mathbb{Q}$: no interval of $\mathbb{Q}$ with more than one point is connected.

Explanations

L1 human

On the real line, the sets in one piece are exactly the intervals. The proof needs the completeness of $\mathbb{R}$, because $\mathbb{Q}$ has gaps where it can be split.