Assuming the Axiom of Choice, $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$: there is $B \subseteq \mathbb{R}$ such that every real is a finite $\mathbb{Q}$-linear combination of elements of $B$ in exactly one way, and each basis vector carries a well-defined $\mathbb{Q}$-linear coefficient map

$\mathbb{Q} = \{\hat q : q \in \mathbb{Q}\}$ is a subfield of $\mathbb{R}$: it contains $\hat 1 = 1$; it is closed under differences and products, since $\hat p - \hat q = \widehat{p-q}$ and $\hat p\,\hat q = \widehat{pq}$; and if $\hat q \ne 0$ then $q \ne 0$ and $\hat q^{-1} = \widehat{q^{-1}}$ lies in it.

$\mathbb{R}$ is a vector space over itself, so restricting the scalars to the subfield $\mathbb{Q}$ makes $\mathbb{R}$ a vector space over $\mathbb{Q}$, with the field addition as vector addition and the field multiplication restricted to $\mathbb{Q} \times \mathbb{R}$ as scalar multiplication.

Claim 1: assuming the Axiom of Choice, that vector space has a basis $B$, a linearly independent subset of $\mathbb{R}$ with $\operatorname{span}(B) = \mathbb{R}$.

Claim 2: since $\operatorname{span}(B) = \mathbb{R}$ and the span of a set is already the set of linear combinations of injective finite lists into it, every real $x$ is $\sum_{i<n}\lambda_i b_i$ with $b : n \to B$ injective and $\lambda : n \to \mathbb{Q}$.

Claim 3: let $b : n \to B$ be injective and put $U := \operatorname{span}(b[n])$, a linear subspace of $\mathbb{R}$. The list $b$ is linearly independent, since $B$ is a linearly independent subset and $b$ is an injective finite list into $B$; its image $b[n]$ spans $U$ by construction, so $b[n]$ is a basis of $U$ and $b$ is an ordered basis of $U$, independence and spanning being the same conditions read in $U$ as in $\mathbb{R}$.

Fix $b_{\star} \in B$ and put $U_{0} := \operatorname{span}\{b_{\star}\} = \{\, \lambda b_{\star} : \lambda \in \mathbb{Q} \,\}$ and $U_{1} := W_{b_{\star}} = \operatorname{span}(B \setminus \{b_{\star}\})$, both linear subspaces of $\mathbb{R}$.

With $b$ and $U$ as in step 4.2, the coordinate theorem applied to the vector space $U$ says that every $x \in U$ is $\sum_{i<n}\lambda_i b_i$ for exactly one $\lambda : n \to \mathbb{Q}$; in particular $\sum_{i<n}\lambda_i b_i = \sum_{i<n}\mu_i b_i$ forces $\lambda = \mu$, which is claim 3.

$U_{0} + U_{1} = \mathbb{R}$. Indeed $U_{0} + U_{1} = \operatorname{span}(U_{0} \cup U_{1})$; the set $B$ is contained in $U_{0} \cup U_{1}$, since $b_{\star} \in U_{0}$ and $B \setminus \{b_{\star}\} \subseteq U_{1}$ by extensiveness of the span, so $\mathbb{R} = \operatorname{span}(B) \subseteq \operatorname{span}(U_{0} \cup U_{1})$ by monotonicity; and $U_{0} \cup U_{1} \subseteq \operatorname{span}(B)$, again by monotonicity, so $\operatorname{span}(U_{0} \cup U_{1}) \subseteq \operatorname{span}(\operatorname{span}(B)) = \operatorname{span}(B) = \mathbb{R}$ by idempotence.

$b_{\star} \ne 0$ and $b_{\star} \notin U_{1}$. If $b_{\star}$ lay in $\operatorname{span}(B \setminus \{b_{\star}\})$ then $B$ would be linearly dependent, contrary to step 3.1; and $0 \in B$ would likewise make $B$ dependent, since $0 \in \operatorname{span}(B \setminus \{0\})$, every span containing the zero vector.

$U_{0} \cap U_{1} = \{0\}$. Let $z \in U_{0} \cap U_{1}$ and write $z = \lambda b_{\star}$ with $\lambda \in \mathbb{Q}$. If $\lambda \ne 0$ then $b_{\star} = \lambda^{-1}z \in U_{1}$, because $U_{1}$ is a linear subspace and $\lambda^{-1} \in \mathbb{Q}$, contradicting step 5.3; so $\lambda = 0$ and $z = 0$.

Hence $\mathbb{R} = U_{0} \oplus U_{1}$: condition (D1) is step 5.2, and condition (D2) is step 6.1, since for the two-member family the sum of the other summands is $U_{1}$ in the one case and $U_{0}$ in the other. By the direct-sum criterion every real $x$ is $u_{0} + u_{1}$ with $u_{0} \in U_{0}$ and $u_{1} \in U_{1}$ in exactly one way.

Writing $u_{0} = \lambda b_{\star}$, the scalar $\lambda \in \mathbb{Q}$ is determined by $u_{0}$, since $b_{\star} \ne 0$; so $\Lambda_{b_{\star}}(x) := \lambda$ is a well-defined map $\mathbb{R} \to \mathbb{Q}$, and $x = \Lambda_{b_{\star}}(x)\,b_{\star} + w$ with $w \in W_{b_{\star}}$ in exactly one way.

$\Lambda_{b_{\star}}$ is additive and $\mathbb{Q}$-homogeneous: if $x = \lambda b_{\star} + w$ and $y = \mu b_{\star} + w'$ with $w, w' \in W_{b_{\star}}$, then $x + y = (\lambda + \mu)b_{\star} + (w + w')$ with $w + w' \in W_{b_{\star}}$, and $qx = (q\lambda)b_{\star} + qw$ with $qw \in W_{b_{\star}}$ for $q \in \mathbb{Q}$, both because $W_{b_{\star}}$ is a linear subspace; uniqueness in step 8.1 then identifies the coefficients.

$\Lambda_{b_{\star}}(b_{\star}) = 1$, from the representation $b_{\star} = 1\cdot b_{\star} + 0$; the range of $\Lambda_{b_{\star}}$ is all of $\mathbb{Q}$, since $\Lambda_{b_{\star}}(q b_{\star}) = q$ for every $q \in \mathbb{Q}$; and $\Lambda_{b_{\star}}(x) = 0$ holds exactly when $x = 0\cdot b_{\star} + w = w \in W_{b_{\star}}$. Claim 4 is proved.

Claim 5: if $W_{b_{\star}} = \{0\}$ then step 8.1 gives $\mathbb{R} = \{\lambda b_{\star} : \lambda \in \mathbb{Q}\}$. That set is the image of $\mathbb{Q}$ under $\lambda \mapsto \lambda b_{\star}$, and $\mathbb{Q}$ is the image of a surjection from $\mathbb{N}$, so composing gives a surjection from $\mathbb{N}$ onto $\mathbb{R}$ and $\mathbb{R}$ would be at most countable, contradicting its uncountability. So $W_{b_{\star}} \ne \{0\}$.