= Solution
<Zorn lemma> states that a nonempty <partially ordered set> in which every chain has an upper bound contains a maximal element. To prove it from the <axiom of choice> and <Hartogs theorem>, suppose no element were maximal. Fix a choice function on the nonempty subsets of the poset. Every chain then has a strict upper bound: choose an upper bound, and, since it is not maximal, choose an element strictly above it. Let $\kappa$ be a Hartogs ordinal which cannot inject into the poset. By <transfinite recursion> for $\alpha<\kappa$, choose a strict upper bound $x_\alpha$ of all previously chosen $x_\beta$, $\beta<\alpha$; for the empty initial chain choose any element. The previous elements form a chain at every stage. The resulting strictly increasing sequence injects $\kappa$ into the poset, contradicting Hartogs. Hence a maximal element exists. \b[Choice is used in fixing the function selecting a strict upper bound at every successor and limit stage]; Hartogs and <transfinite recursion> themselves do not supply those selections.
Apply <Zorn lemma> to the linearly independent subsets of $\mathbb R$ over $\mathbb Q$, ordered by inclusion. A chain's union is linearly independent because any finite linear relation is contained in one member of the chain. A maximal independent set $B$ must span $\mathbb R$, since otherwise one could append a <vector> outside its span. Thus \b[$B$ is a rational-vector-space basis of $\mathbb R$].
It is infinite: a finite-dimensional rational <vector space> is countable, whereas $\mathbb R$ is not. If its cardinal is $\kappa$, finite rational linear combinations show $|\mathbb R|=\max(\aleph_0,\kappa)$; uncountability forces $\kappa=|\mathbb R|$. A basis of $\mathbb R^2$ is $(B\times\{0\})\cup(\{0\}\times B)$, where these expressions mean the <vectors> $(b,0)$ and $(0,b)$. Its cardinal is $\kappa+\kappa=\kappa$, using the infinite-cardinal arithmetic available under choice. A bijection between the two bases extends uniquely by finite rational linear combinations to a linear bijection. Therefore
$$
\boxed{\mathbb R\cong_{\mathbb Q}\mathbb R^2.}
$$
This is an algebraic isomorphism, with no assertion of continuity or real-linearity.
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