Taras Banakh writes in the original question that *by transfinite induction of length $\mathfrak c$, one can construct a dense $\mathbb Q$-linear subspace $L$ of the Euclidean plane $\mathbb R^2$ such that $|\{x\in L:\|x\|=r\}|=2$ for every positive real number $r$*. This is indeed rather straightforward to do; it’s a just-do-it proof.

First, observe that if $L$ has the properties above, the translation invariance of $L$ implies that it is a Banakh space, but $L$ is not isometric to $\mathbb R$ as it is not complete. Thus, it answers the question in the negative.

To construct $L$, let us fix an enumeration $\mathbb R_{>0}=\{r_\alpha:\alpha<\mathfrak c\}$. We will build by transfinite recursion a sequence $X=\{x_\alpha:\alpha<\mathfrak c\}\subseteq\mathbb R^2$ such that $\|x_\alpha\|=r_\alpha$, and such that $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L$, where $L$ is the $\mathbb Q$-linear span $\mathbb QX$. Let $\alpha<\mathfrak c$, and assume that $\{x_\beta:\beta<\alpha\}$ has been already defined such that $\|x_\beta\|=r_\beta$, and $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L_\alpha=\mathbb Q\{x_\beta:\beta<\alpha\}$; we will define $x_\alpha$.

If $\|a\|=r_\alpha$ for some $a\in L_\alpha$, we can define $x_\alpha$ as one of the two such elements $a$. Thus, we may assume $L_\alpha$ is disjoint from the circle $C_\alpha=\{x\in\mathbb R^2:\|x\|=r_\alpha\}$.

We pick $x_\alpha$ as a suitable element $x\in C_\alpha$. We need to ensure that for each $a,b\in L_\alpha$ and $q,r\in\mathbb Q$ such that $(a,q)\ne\pm(b,r)$, $\|a+qx\|\ne\|b+rx\|$. This is a collection of $|\alpha|+\aleph_0<\mathfrak c$ possible obstructions; we will show that each obstruction set is a line or a circle different from $C_\alpha$, and therefore intersects $C_\alpha$ in at most $2$ points. Thus, only $|\alpha|+\aleph_0<\mathfrak c$ points of $C_\alpha$ violate some obstruction, and we can choose $x_\alpha$ as any of the remaining $\mathfrak c$ points. (Moreover, if $\alpha=1$, we can make sure $x_1$ is not a scalar multiple of $x_0$, thus the resulting set $L$ will not be contained in a line; being a $\mathbb Q$-linear set, this will make it dense in $\mathbb R^2$.)

The obstructions $\|a+qx\|=\|b+rx\|$ are of the following kind:

$q=r=0$: Then $b\ne\pm a$, thus $\|a\|\ne\|b\|$ by the induction hypothesis, i.e., the obstruction set is empty.

$q=\pm r\ne0$: Scaling everything by $q^{-1}$, and possibly negating $(b,r)$, we may assume $q=r=1$ and $a\ne b$. Then the obstruction set $\{x:\|x+a\|=\|x+b\|\}$ is a line (the perpendicular bisector of $-a$ and $-b$).

$q\ne\pm r$: If you do the algebra, it is easy to see that the obstruction set $\{x:\|qx+a\|=\|rx+b\|\}$ is a circle. Moreover, it contains the point $x=(a-b)/(r-q)\in L_\alpha$, hence the circle is different from $C_\alpha$.

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