$\mathbb{R}$ as a vector space over $\mathbb{Q}$ has a basis, and every such basis is infinite; the existence proof exhibits none

Claim 1. ${\mathbb{Q}}_{\mathbb{R}}=\iota [\mathbb{Q}]$ is a subfield of $\mathbb{R}$: it contains $\iota (1)=1$; for $p,q\in \mathbb{Q}$ it contains $\iota (p)-\iota (q)=\iota (p-q)$ and $\iota (p)\iota (q)=\iota (pq)$; and if $\iota (q)\mathrm{\ne }0$ then $q\mathrm{\ne }0$, since $\iota (0)=0$, so $\iota (q{)}^{-1}=\iota (q^{-1})\in {\mathbb{Q}}_{\mathbb{R}}$. Since $\mathbb{R}$ is a vector space over itself, restriction of scalars makes it a vector space over ${\mathbb{Q}}_{\mathbb{R}}$, with the field multiplication restricted to ${\mathbb{Q}}_{\mathbb{R}}\times \mathbb{R}$. The operation $(q,x)\mapsto \iota (q)x$ is a map $\mathbb{Q}\times \mathbb{R}\to \mathbb{R}$, and it satisfies (V2) to (V5) because $\iota$ preserves sums and products and $\iota (1)=1$, while (V1) is the abelian group $(\mathbb{R},+,0)$; so it makes $\mathbb{R}$ a vector space over $\mathbb{Q}$.

${\mathbb{Q}}_{\mathbb{R}}$ is at most countable: $\iota$ is injective with image ${\mathbb{Q}}_{\mathbb{R}}$, hence a bijection $\mathbb{Q}\to {\mathbb{Q}}_{\mathbb{R}}$, and $\mathbb{Q}\approx \mathbb{N}$; composing bijections gives ${\mathbb{Q}}_{\mathbb{R}}\approx \mathbb{N}$.

If $K$ is at most countable then so is $K^n$, the set of functions $n\to K$, for every $n\in \mathbb{N}$. By induction on $n$: at $n=0$ the set $K^0$ has exactly one element, the empty function, so it is finite; and the map $K^{\sigma (n)}\to K^n\times K$ sending $f$ to the pair consisting of its restriction to $n$ and its value at $n$ is a bijection, since $\sigma (n)=n\cup \{n\}$ and $n\mathrm{\notin }n$, so a function on $\sigma (n)$ is determined by, and may be assembled from, those two data. Hence $K^{\sigma (n)}\approx K^n\times K$, which is at most countable by the inductive hypothesis and the product theorem, and countability transfers along the bijection.

Claim 4. For a list $v:n\to \mathbb{R}$ and scalars $\lambda :n\to \mathbb{Q}$, the vector ${\sum }_{i<n}{\lambda }_i\cdot v_i$ computed in the $\mathbb{Q}$-structure is by definition ${\sum }_{i<n}\iota ({\lambda }_i)v_i$, computed in the ${\mathbb{Q}}_{\mathbb{R}}$-structure; the two structures have the same underlying set, the same addition and the same zero, so their finite sums agree. Since $\iota$ is a bijection $\mathbb{Q}\to {\mathbb{Q}}_{\mathbb{R}}$, the scalar lists $\lambda :n\to \mathbb{Q}$ and $\iota \circ \lambda :n\to {\mathbb{Q}}_{\mathbb{R}}$ correspond bijectively, and ${\lambda }_i=0$ for all $i$ exactly when $\iota ({\lambda }_i)=0$ for all $i$. So a vanishing combination exists on one side exactly when it does on the other, and likewise for representations of an arbitrary vector; hence the two structures have the same linearly independent subsets, the same spans and the same bases.

Claim 2. $\mathbb{R}$ is a vector space over ${\mathbb{Q}}_{\mathbb{R}}$ by step 1.1, and every vector space has a basis, so a basis $B$ of $\mathbb{R}$ over ${\mathbb{Q}}_{\mathbb{R}}$ exists.

Claim 3. Suppose some basis $B$ of $\mathbb{R}$ over ${\mathbb{Q}}_{\mathbb{R}}$ were finite, say $B\approx n$. A bijection $n\to B$ is an injective list whose image is a basis, hence an ordered basis, so every $x\in \mathbb{R}$ is ${\sum }_{i<n}{\lambda }_i{b}_i$ for exactly one $\lambda :n\to {\mathbb{Q}}_{\mathbb{R}}$. The resulting map $\mathrm{\Phi }:\mathbb{R}\to ({\mathbb{Q}}_{\mathbb{R}}{)}^n$, sending $x$ to that $\lambda$, is injective, since $\mathrm{\Phi }(x)=\mathrm{\Phi }(y)$ makes $x$ and $y$ the same sum. By steps 1.2 and 1.3 the set $({\mathbb{Q}}_{\mathbb{R}}{)}^n$ is at most countable, hence so is its subset $\mathrm{\Phi }[\mathbb{R}]$; and $\mathrm{\Phi }$ is a bijection $\mathbb{R}\to \mathrm{\Phi }[\mathbb{R}]$, so $\mathbb{R}$ is at most countable, contradicting the uncountability of $\mathbb{R}$. So no basis of $\mathbb{R}$ over ${\mathbb{Q}}_{\mathbb{R}}$ is finite, and $\mathbb{R}$ is infinite-dimensional over ${\mathbb{Q}}_{\mathbb{R}}$.

Claim 1 is step 1.1, claim 2 is step 2.1, claim 3 is step 2.2, and claim 4 is step 1.4; by claim 4 the last two transfer to $\mathbb{R}$ as a $\mathbb{Q}$-vector space.