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 $c \ne a$, 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 $c \ne b$, 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 $b \notin U$ 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 $U \cap V = \varnothing$.

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 $C$.