$\mathbb R$ is not isometric with $\mathbb R^2$

In $\mathbb R$ there do not exist three distinct points such that the distance between each pair is equal to $1$.

On the other hand, in $\mathbb R^2$ there do exist equilateral triangles.

An isometry between $\mathbb R$ and $\mathbb R^2$ would imply a contradiction. Consider that any one point separates $\mathbb R$ , but does not separate/disconnect $\mathbb R^2$. If h:$\mathbb R \rightarrow \mathbb R^2$ were a homeomorphism between the two, then, for any $x$ in $\mathbb R$ , h':$\mathbb R-{x}\rightarrow \mathbb R^2-h(x)$ would also be a homeomorphism. But this is not possible, since $\mathbb R-{x}$ is disconnected , but $\mathbb R^2-h(x)$ is not.

Maybe to be more precise, if there was a continuous bijection h (an isometry) between $\mathbb R$ and $\mathbb R^2$ , then the following contradiction would result:

Consider the restriction h' of h to $[-1,1]$ . By compactness of [-1,1], and by $\mathbb R^2$ being Hausdorff (so that its subspace h'([-1,1]) is Hausdorff ), we have a continuous bijection between compact and Hausdorff, so that $h'([-1,1])\rightarrow h'([-1,1])$ is a homeomorphism. By connectedness of [-1,1] and since h' is injective (and h is a continuous bijection into $\mathbb R^2$), the image contains an open ball. Now , since h' is a homeomorphism, it sends its interior (-1,1) into the interior of the image, which is connected , but we have the issue of the cutsets.