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

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

Claim 2: since $\mathrm{span}(B)=\mathbb{R}$ 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 $U:=\mathrm{span}(b[n])$, 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 $U_0:=\mathrm{span}\{b_{\star }\}=\{\lambda b_{\star }:\lambda \in \mathbb{Q}\}$ and $U_1:=W_{b_{\star }}=\mathrm{span}(B\setminus \{b_{\star }\})$, 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 $U_0+U_1=\mathrm{span}(U_0\cup U_1)$; 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 $\mathbb{R}=\mathrm{span}(B)\subseteq \mathrm{span}(U_0\cup U_1)$ by monotonicity; and $U_0\cup U_1\subseteq \mathrm{span}(B)$, again by monotonicity, so $\mathrm{span}(U_0\cup U_1)\subseteq \mathrm{span}(\mathrm{span}(B))=\mathrm{span}(B)=\mathbb{R}$ by idempotence.

$b_{\star }\mathrm{\ne }0$ and $b_{\star }\mathrm{\notin }{U}_1$. If $b_{\star }$ lay in $\mathrm{span}(B\setminus \{b_{\star }\})$ then $B$ would be linearly dependent, contrary to step 3.1; and $0\in B$ would likewise make $B$ dependent, since $0\in \mathrm{span}(B\setminus \{0\})$, 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 $\lambda \mathrm{\ne }0$ 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 $b_{\star }\mathrm{\ne }0$; so ${\mathrm{\Lambda }}_{b_{\star }}(x):=\lambda$ is a well-defined map $\mathbb{R}\to \mathbb{Q}$, and $x={\mathrm{\Lambda }}_{b_{\star }}(x)b_{\star }+w$ with $w\in W_{b_{\star }}$ in exactly one way.

${\mathrm{\Lambda }}_{b_{\star }}$ is additive and $\mathbb{Q}$-homogeneous: if $x=\lambda b_{\star }+w$ and $y=\mu b_{\star }+w^{\mathrm{\prime }}$ with $w,w^{\mathrm{\prime }}\in W_{b_{\star }}$, then $x+y=(\lambda +\mu )b_{\star }+(w+w^{\mathrm{\prime }})$ with $w+w^{\mathrm{\prime }}\in W_{b_{\star }}$, 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.

${\mathrm{\Lambda }}_{b_{\star }}(b_{\star })=1$, from the representation $b_{\star }=1\cdot b_{\star }+0$; the range of ${\mathrm{\Lambda }}_{b_{\star }}$ is all of $\mathbb{Q}$, since ${\mathrm{\Lambda }}_{b_{\star }}(q{b}_{\star })=q$ for every $q\in \mathbb{Q}$; and ${\mathrm{\Lambda }}_{b_{\star }}(x)=0$ 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 $W_{b_{\star }}\mathrm{\ne }\{0\}$.