$\mathbb{R}$ 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 $\iota(q) \ne 0$ then $q \ne 0$, 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 $n \notin n$, 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 $\Phi : \mathbb{R} \to (\mathbb{Q}_{\mathbb{R}})^{n}$, sending $x$ to that $\lambda$, is injective, since $\Phi(x) = \Phi(y)$ 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 $\Phi[\mathbb{R}]$; and $\Phi$ is a bijection $\mathbb{R} \to \Phi[\mathbb{R}]$, 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.