The long ray is a linear continuum, hence connected; every one of its at most countable subsets is bounded above, assuming countable choice

$R$ has at least two elements, namely $(0,0)$ and $(0,1\mathrm{/}2)$, which differ and satisfy $(0,0)<(0,1\mathrm{/}2)$ by [A1] and [A4].

$R$ is order-dense. Let $(\alpha ,s)<(\beta ,t)$. If $\alpha =\beta$ then $s<t$ and $(\alpha ,(s+t)\mathrm{/}2)$ lies strictly between, by [A4]. If $\alpha <\beta$ then $s<(s+1)\mathrm{/}2<1$ by [A4], so $(\alpha ,(s+1)\mathrm{/}2)\in R$ lies strictly above $(\alpha ,s)$ and strictly below $(\beta ,t)$, its first coordinate being $\alpha <\beta$.

Let $S\subseteq R$ be nonempty with an upper bound $({\beta }_0,t_0)$, and put $A:=\{\alpha \in {\omega }_1:(\alpha ,s)\in S\text{ for some }s\}$, a nonempty set of ordinals with $\alpha \le {\beta }_0$ for every $\alpha \in A$; so $\gamma :=\bigcup A$ is an ordinal with $\gamma \le {\beta }_0$, hence $\gamma \in {\omega }_1$ by [A2] and [A3].

For claim 3 let $D\subseteq R$ be at most countable. If $D=\mathrm{\varnothing }$ then $0_R$ is an upper bound of $D$ by [A1] and there is nothing more to prove, so assume $D\mathrm{\ne }\mathrm{\varnothing }$ and let $A_D:=\{\alpha :(\alpha ,s)\in D\text{ for some }s\}$, a nonempty subset of ${\omega }_1$.

Suppose first $\gamma \in A$, and put $T:=\{s\in [0,1):(\gamma ,s)\in S\}$, which is nonempty and bounded above by $1$; let $u:=\sup T$ in $\mathbb{R}$, which exists and is unique by [A4], with $0\le u\le 1$.

Suppose instead $\gamma \mathrm{\notin }A$; then every $\alpha \in A$ satisfies $\alpha <\gamma$, since $\alpha \le \gamma$ by [A2] and $\alpha \mathrm{\ne }\gamma$.

$A_D$ is at most countable: by [A7] there is a surjection $f:\mathbb{N}\to D$, and composing it with the first-coordinate map gives a surjection $\mathbb{N}\to A_D$, so [A7] applies again.

In the case of step 2.1 with $u<1$, the element $(\gamma ,u)\in R$ is the least upper bound of $S$: it bounds $S$, since $(\alpha ,s)\in S$ has $\alpha \le \gamma$ and, when $\alpha =\gamma$, $s\in T$ so $s\le u$; and any upper bound $(\beta ,t)$ of $S$ has $\beta \ge \gamma$, because $S$ contains an element with first coordinate $\gamma$, and if $\beta =\gamma$ then $t$ bounds $T$ so $t\ge u$.

In the case of step 2.1 with $u=1$, the element $({\gamma }^{+},0)\in R$ is the least upper bound of $S$: it bounds $S$, since every $(\alpha ,s)\in S$ has $\alpha \le \gamma <{\gamma }^{+}$; and an upper bound $(\beta ,t)$ cannot have $\beta <\gamma$, $S$ containing an element with first coordinate $\gamma$, nor $\beta =\gamma$, since then $t$ would bound $T$ and give $t\ge u=1$ against $t<1$; so $\beta >\gamma$, that is $\beta \ge {\gamma }^{+}$ by [A2], and $(\beta ,t)\ge ({\gamma }^{+},0)$. Here ${\gamma }^{+}\in {\omega }_1$ by [A3].

In the case of step 2.2, the element $(\gamma ,0)\in R$ is the least upper bound of $S$: it bounds $S$, since every $(\alpha ,s)\in S$ has $\alpha <\gamma$; and if an upper bound $(\beta ,t)$ had $\beta <\gamma =\bigcup A$ then $\beta$ would not bound $A$ by [A2], so some $\alpha \in A$ has $\beta <\alpha$ and the corresponding element of $S$ exceeds $(\beta ,t)$ — impossible; so $\beta \ge \gamma$ and $(\beta ,t)\ge (\gamma ,0)$.

By [A6] the ordinal $\mu :=\sup A_D$ lies in ${\omega }_1$ and satisfies $\alpha \le \mu$ for every $\alpha \in A_D$; this is the one step at which ${\mathrm{A}\mathrm{C}}_{\omega }$ is spent.

Steps 1.3, 2.1, 2.2, 3.1, 3.2 and 3.3 exhaust the cases and give a least upper bound in each, so $R$ has the least upper bound property; with steps 1.1 and 1.2 this makes $R$ a linear continuum by [A5]. This is claim 1.

Claim 2 follows: $R$ is connected and every order-convex subset of $R$ is connected by [A5], and each initial segment $[0_R,x]$ is order-convex, being defined by an inequality closed under passing to intermediate points.

Then ${\mu }^{+}\in {\omega }_1$ by [A3], and $({\mu }^{+},0)\in R$ is an upper bound of $D$: every $(\alpha ,s)\in D$ has $\alpha \in A_D$, hence $\alpha \le \mu <{\mu }^{+}$ by [A2] and step 3.4, so $(\alpha ,s)<({\mu }^{+},0)$. This is claim 3.