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You cannot prove that $\mathbb{R}$ and $\mathbb{R}^{2}$ are isomorphic in $ZF$. To see this, note that the map $(x,y) \mapsto (x,0)$ is a nontrivial non-surjective additive endomorphism of $\mathbb{R}^{2}$. Assuming suitable large cardinals, the only additive endomorphisms of $\mathbb{R}$ in $L(\mathbb{R})$ are the maps $x \mapsto rx$ for some $r \in \mathbb{R}$. (This is because, assuming suitable large cardinals, every map $f: \mathbb{R} \to \mathbb{R}$ is measurable in $L(\mathbb{R})$.) Thus $\mathbb{R} \not \cong \mathbb{R}^{2}$ in $L(\mathbb{R})$.