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 $w\mathrm{\notin }E$; then $w\mathrm{\ne }x$ and $w\mathrm{\ne }z$, so $x<w<z$. Put $A:=E\cap (-\mathrm{\infty },w)$ and $B:=E\cap (w,\mathrm{\infty })$; 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 $p\mathrm{\ne }q$, 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, $\sup S\in \bar{S}$: for every real $\epsilon >0$ the fact [L5] supplies $s\in S$ with $\sup S-\epsilon <s\le \sup S$, so $\mathrm{\mid }s-\sup S\mathrm{\mid }<\epsilon$ and $s\in N_{\epsilon }(\sup S)\cap S$; thus every neighbourhood of $\sup S$ meets $S$, and [L2] gives $\sup S\in \bar{S}$.

In the situation of step 1.1 the pair $(A,B)$ is a disconnection: $(-\mathrm{\infty },w]$ is a closed set containing $A$, so $\bar{A}\subseteq (-\mathrm{\infty },w]$ by [L2], whence $\bar{A}\cap B\subseteq (-\mathrm{\infty },w]\cap (w,\mathrm{\infty })=\mathrm{\varnothing }$; symmetrically $\bar{B}\subseteq [w,\mathrm{\infty })$ and $A\cap \bar{B}=\mathrm{\varnothing }$. 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 $\bar{S}\subseteq \bar{A}$, and $c\in \bar{S}$ by step 1.3, so $c\in \bar{A}$ and hence $c\mathrm{\notin }B$ because $\bar{A}\cap B=\mathrm{\varnothing }$; 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 $v\mathrm{\notin }A$, for otherwise $v\in A\cap [p,q]=S$ would force $v\le c$.

$c\in \bar{B}$, which is impossible: given a real $\epsilon >0$, put $t:=\min \{\epsilon \cdot 2^{-1},(q-c)\cdot 2^{-1}\}$, 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 $\mathrm{\mid }v-c\mathrm{\mid }=t\le \epsilon \cdot 2^{-1}<\epsilon$, so $v\in N_{\epsilon }(c)\cap B$. Hence every neighbourhood of $c$ meets $B$ and $c\in \bar{B}$ by [L2]; but $c\in A$ by step 3.1 and $A\cap \bar{B}=\mathrm{\varnothing }$ 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.