Is there a topological way to see that $\mathbb R$ and $\mathbb R^2$ have equal cardinality? I know the "interleaving" proof sending the tuple $(0.x_1x_2\dots,0.y_1y_2\dots)\mapsto0.x_1y_1x_2y_2\dots$, showing $(0,1)\cong(0,1)^2$, which can be extended to $\mathbb R\cong\mathbb R^2$. But this relies on a particular representation of real numbers, and the bijection behaves quite poorly from a topological point of view.
Since the reals are at their heart a topological construction (the completion of the rationals), as well as $\mathbb R$ and $\mathbb R^2$ both having very clear geometric (and thus topological) interpretations, I wonder wether we can see $\mathbb R\cong\mathbb R^2$ due to topological reasons.
Alternatively, I can imagine that the fact that $X\cong X^2$ is a more general fact which does not depend on topological facts about the reals in the first place. Can we show that every infinite set $X$ (infinite in whatever sense you want, as long as the reals are covered by it) satisfies $X\cong X^2$?
 A: As mentioned in the comments, there are continuous surjections in either direction between $\mathbb{R}$ and $\mathbb{R}^2$ and of course $\mathbb{R}$ embeds in $\mathbb{R}^2$ as a subspace. On the other hand there is no continuous injection $\mathbb{R}^2 \to \mathbb{R}$, the image has to be an interval, and removing a point of $\mathbb{R}^2$ that maps to an interior point of this interval leaves the domain connected but disconnects the image. This restricts what forms of topological proof of the bijection are possible.
Here is a generalization of the interleaving argument: since $\mathbb{N} \cong \mathbb{N} \sqcup \mathbb{N}$, for a set of the form $X = Y^\mathbb{N}$ we have $X \cong X^2$. This gives us the fact about $\mathbb{R}$ after deciding on a bijection $\mathbb{R} \cong \{0, \dotsc, 9\}^{\mathbb{N}}$. You could also replace $\mathbb{N}$ with other infinite sets, though the usual bijection $\mathbb{N} \cong \mathbb{N} \sqcup \mathbb{N}$ is particularly easy to write down explicitly.
As mentioned in the comments again, you need to use some choice to show for general infinite sets $X$ that $X \cong X^2$.
