A linear continuum is connected in its order topology, and so is every order-convex subset of it

Suppose, for contradiction, that $(U,V)$ is a separation of $L$: both open and nonempty, disjoint, with $U\cup V=L$.

Fix $a\in U$ and $b\in V$. They are distinct, $U$ and $V$ being disjoint, so by [A1] one is below the other; relabelling $U$ and $V$ if necessary, which is legitimate because the hypothesis of step 1.1 is symmetric in them, assume $a<b$.

Put $S:=\{t\in U:a\le t\le b\}$. It is nonempty, containing $a$, and $b$ is an upper bound of it, so $c:=\sup S$ exists by [A2] and satisfies $a\le c\le b$, since $a\in S$ and $b$ is an upper bound.

Suppose $c\in V$. Then $c\mathrm{\ne }a$, since $a\in U$ and the two sets are disjoint, so $a<c$ by [A1] and $a\le c$. By [A3] there is a basic open $W$ with $c\in W\subseteq V$; $W$ is neither $L$ nor a set $L_{<q}$, since either would contain $a$, giving $a\in V$. So $W$ is $L_{>p}$ or $(p,q)$ with $p<c$, and in both cases $(p,c]\subseteq W\subseteq V$.

Suppose instead $c\in U$. Then $c\mathrm{\ne }b$, so $c<b$ by [A1] and $c\le b$. By [A3] there is a basic open $W$ with $c\in W\subseteq U$; $W$ is neither $L$ nor a set $L_{>p}$, since either would contain $b$, giving $b\in U$. So $W$ is $L_{<q}$ or $(p,q)$ with $c<q$, and in both cases $[c,q)\subseteq W\subseteq U$; moreover $q\le b$, since $b\mathrm{\notin }U$ and $c<b$ would otherwise put $b$ in $[c,q)$.

In the case of step 4.1, $p<c=\sup S$, so $p$ is not an upper bound of $S$ by [A2] and there is $s\in S$ with $p<s\le c$; then $s\in (p,c]\subseteq V$ and $s\in U$, contradicting $U\cap V=\mathrm{\varnothing }$.

In the case of step 4.2, order-density gives $z$ with $c<z<q$ by [A2]; then $z\in [c,q)\subseteq U$, and $a\le c<z<q\le b$, so $z\in S$ while $z>c=\sup S$, contradicting that $\sup S$ is an upper bound of $S$.

By step 1.1 the point $c$ lies in $U\cup V=L$, so one of the two cases applies, and each is contradictory by steps 5.1 and 5.2. Hence no separation of $L$ exists and $L$ is connected; this is claim 1.

For claim 2 let $C\subseteq L$ be order-convex. If $C$ has at most one element it is connected by [A4]. Otherwise $C$ carries the order topology of its restricted order by [A5], and $C$ is itself a linear continuum: it has at least two elements; it is order-dense, because for $x<y$ in $C$ the element $z$ with $x<z<y$ given by [A2] lies in $C$ by order-convexity; and it has the least upper bound property, because a nonempty $S\subseteq C$ with an upper bound $u\in C$ has $\sup S$ in $L$ by [A2], and $s\le \sup S\le u$ for any $s\in S$ puts $\sup S$ in $C$ by order-convexity, where it is again the least upper bound. So claim 1 applies to $C$.