Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, namely its order components

The hull satisfies $x, y \in H(x,y)$, $H(x,y) = H(y,x)$ and $H(x,x) = \{x\}$, and for all $x, y, z$ one has $H(x,z) \subseteq H(x,y) \cup H(y,z)$: given $w \in H(x,z)$, either $x \le w \le z$, in which case $w \le y$ puts $w$ in $H(x,y)$ and $y < w$ puts $w$ in $H(y,z)$, or $z \le w \le x$, in which case $w \le y$ puts $w$ in $H(y,z)$ and $y < w$ puts $w$ in $H(x,y)$. Hence $\sim$ is reflexive on $U$ (as $H(x,x) = \{x\} \subseteq U$), symmetric, and transitive.

Let $C \subseteq \mathbb{R}$ be nonempty, open and order-convex, and let $u \in C$; fix $\varepsilon > 0$ with $N_\varepsilon(u) \subseteq C$. Then $u - \varepsilon/2$ and $u + \varepsilon/2$ lie in $N_\varepsilon(u) \subseteq C$, so $u$ is neither an upper bound nor a lower bound of $C$.

For $x \in U$ put $C_x := \{\, y \in U : H(x,y) \subseteq U \,\}$, the equivalence class of $x$, and let $\mathcal{C} := \{\, C_x : x \in U \,\}$.

Each $C_x$ is nonempty because $x \in C_x$; two classes of an equivalence relation are equal or disjoint; and every $x \in U$ lies in $C_x$, so $U = \bigcup \mathcal{C}$. This is claim 1.

Each $C_x$ is order-convex: let $u, v \in C_x$ and $u \le w \le v$. From $u \sim x$ and $x \sim v$ we get $u \sim v$, so $H(u,v) \subseteq U$; since $w \in H(u,v)$ we get $w \in U$, and $H(u,w) \subseteq H(u,v) \subseteq U$ because every $t$ with $u \le t \le w$ satisfies $u \le t \le v$, so $u \sim w$ and hence $w \in C_u = C_x$.

Each $C_x$ is open: let $u \in C_x \subseteq U$ and fix $\varepsilon > 0$ with $N_\varepsilon(u) \subseteq U$. For $y \in N_\varepsilon(u)$ the hull $H(u,y)$ is contained in the order-convex set $N_\varepsilon(u)$, hence in $U$, so $u \sim y$ and $y \in C_u = C_x$; therefore $N_\varepsilon(u) \subseteq C_x$.

Let $C$ be nonempty, open and order-convex and bounded both above and below; then $a := \inf C$ and $b := \sup C$ exist by [L4]. Every $u \in C$ satisfies $a \le u \le b$, and $u$ is neither an upper nor a lower bound of $C$, so $u \ne a$ and $u \ne b$, giving $a < u < b$; in particular $a < b$ and $C \subseteq (a,b)$. Conversely let $a < w < b$: by [L5] with $\varepsilon = b - w$ there is $v \in C$ with $w < v$, and with $\varepsilon = w - a$ there is $t \in C$ with $t < w$, so $t \le w \le v$ and order-convexity gives $w \in C$. Hence $C = (a,b)$.

Let $C$ be nonempty, open and order-convex. If $C$ is bounded below and not above, put $a := \inf C$; as in the bounded case every $u \in C$ satisfies $a < u$, and for $w > a$ the fact [L5] supplies $t \in C$ with $t < w$ while [L6] supplies $v \in C$ with $w < v$, so $w \in C$ by order-convexity; hence $C = (a,\infty)$. Symmetrically, if $C$ is bounded above and not below then $C = (-\infty, b)$ with $b := \sup C$. If $C$ is bounded neither above nor below then for every $w$ the fact [L6] supplies $t, v \in C$ with $t < w < v$, so $w \in C$ and $C = \mathbb{R}$.

Every member of $\mathcal{C}$ is nonempty, open and order-convex by steps 2.1, 2.2 and 2.3, and it is bounded above or not and bounded below or not, so steps 2.4 and 2.5 exhibit it as an interval of one of the four open forms; this is claim 2.

Every member $C$ of $\mathcal{C}$ contains an element of $\mathbb{Q}_{\mathbb{R}}$: pick $u \in C$ and, by openness, $\varepsilon > 0$ with $N_\varepsilon(u) \subseteq C$; since $u - \varepsilon < u + \varepsilon$, the fact [L7] supplies $\hat q$ with $u - \varepsilon < \hat q < u + \varepsilon$, and $N_\varepsilon(u) = (u - \varepsilon, u + \varepsilon)$ by [L2], so $\hat q \in C$.

By [L8] fix a bijection $\beta : \mathbb{N} \to \mathbb{Q}$; then $e := \iota \circ \beta$, where $\iota(q) = \hat q$, is a bijection from $\mathbb{N}$ onto $\mathbb{Q}_{\mathbb{R}}$ by [L7] and [L8]. For $C \in \mathcal{C}$ the set $\{\, n \in \mathbb{N} : e(n) \in C \,\}$ is nonempty by step 3.2, so $\Phi(C) := \min \{\, n \in \mathbb{N} : e(n) \in C \,\}$ is defined by [L9] and no selection is made; and $\Phi$ is injective, since $e(\Phi(C)) \in C$ and distinct members of $\mathcal{C}$ are disjoint by step 2.1.

Hence $\mathcal{C}$ is in bijection with $\Phi[\mathcal{C}] \subseteq \mathbb{N}$, and a subset of $\mathbb{N}$ is at most countable, so $\mathcal{C}$ is at most countable; this is claim 3.

The family $\mathcal{C}$ constructed in step 1.3 therefore consists of pairwise disjoint nonempty open intervals whose union is $U$, and it is at most countable, which is exactly the assertion.