A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval

Suppose $E$ is not order-convex: there are $x, z \in E$ and $w \in \mathbb{R}$ with $x \le w \le z$ and $w \notin E$; then $w \ne x$ and $w \ne z$, so $x < w < z$. Put $A := E \cap (-\infty,w)$ and $B := E \cap (w,\infty)$; then $x \in A$ and $z \in B$, so both are nonempty, and $A \cup B = E$ because no element of $E$ equals $w$.

Suppose instead that $E$ is order-convex and that $(A,B)$ is a disconnection of $E$; fix $p \in A$ and $q \in B$. Separated sets are disjoint by [L1], so $p \ne q$, and interchanging the names $A$ and $B$ if necessary, which is legitimate because the hypotheses on the pair are symmetric, we may assume $p < q$.

For a nonempty $S \subseteq \mathbb{R}$ bounded above, $\sup S \in \overline{S}$: for every real $\varepsilon > 0$ the fact [L5] supplies $s \in S$ with $\sup S - \varepsilon < s \le \sup S$, so $|s - \sup S| < \varepsilon$ and $s \in N_\varepsilon(\sup S) \cap S$; thus every neighbourhood of $\sup S$ meets $S$, and [L2] gives $\sup S \in \overline{S}$.

In the situation of step 1.1 the pair $(A,B)$ is a disconnection: $(-\infty,w]$ is a closed set containing $A$, so $\overline{A} \subseteq (-\infty,w]$ by [L2], whence $\overline{A} \cap B \subseteq (-\infty,w] \cap (w,\infty) = \varnothing$; symmetrically $\overline{B} \subseteq [w,\infty)$ and $A \cap \overline{B} = \varnothing$. So $A$ and $B$ are separated, nonempty, and their union is $E$, and $E$ is disconnected.

In the situation of step 1.2 put $S := A \cap [p,q]$; it is nonempty because $p \in A$ and $p \le p \le q$, and it is bounded above by $q$, so $c := \sup S$ exists by [L4], and $p \le c \le q$ since $p \in S$ and $q$ is an upper bound.

$c \in A$: from $S \subseteq A$ and [L2] we get $\overline{S} \subseteq \overline{A}$, and $c \in \overline{S}$ by step 1.3, so $c \in \overline{A}$ and hence $c \notin B$ because $\overline{A} \cap B = \varnothing$; on the other hand $p \le c \le q$ with $p, q \in E$ and $E$ order-convex gives $c \in E = A \cup B$, so $c \in A$.

$c < q$, since $c \in A$ and $q \in B$ are distinct by [L1] while $c \le q$; and every $v$ with $c < v \le q$ lies in $B$: such a $v$ satisfies $p \le c < v \le q$, so $v \in E$ by order-convexity, and $v \notin A$, for otherwise $v \in A \cap [p,q] = S$ would force $v \le c$.

$c \in \overline{B}$, which is impossible: given a real $\varepsilon > 0$, put $t := \min\{\varepsilon \cdot 2^{-1},\ (q - c) \cdot 2^{-1}\}$, a positive real by [L7] and [L8] since $q - c > 0$, and $v := c + t$; then $c < v$ and $v \le c + (q-c) \cdot 2^{-1} < q$, so $v \in B$ by step 4.1, while $|v - c| = t \le \varepsilon \cdot 2^{-1} < \varepsilon$, so $v \in N_\varepsilon(c) \cap B$. Hence every neighbourhood of $c$ meets $B$ and $c \in \overline{B}$ by [L2]; but $c \in A$ by step 3.1 and $A \cap \overline{B} = \varnothing$ by [L1]. So the assumed disconnection cannot exist and an order-convex $E$ is connected.

Step 2.1 shows that a set which is not order-convex is disconnected, hence a connected set is order-convex; step 5.1 shows that an order-convex set admits no disconnection, hence is connected. The two together are the asserted equivalence.