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 $(0,1/2)$, which differ and satisfy $(0,0) < (0,1/2)$ by [A1] and [A4].

$R$ is order-dense. Let $(\alpha,s) < (\beta,t)$. If $\alpha = \beta$ then $s < t$ and $(\alpha, (s+t)/2)$ lies strictly between, by [A4]. If $\alpha < \beta$ then $s < (s+1)/2 < 1$ by [A4], so $(\alpha, (s+1)/2) \in R$ 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 $D = \varnothing$ then $0_R$ is an upper bound of $D$ by [A1] and there is nothing more to prove, so assume $D \ne \varnothing$ 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 $\gamma \notin A$; then every $\alpha \in A$ satisfies $\alpha < \gamma$, since $\alpha \le \gamma$ by [A2] and $\alpha \ne \gamma$.

$A_D$ 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 $\mathrm{AC}_\omega$ 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.