Is $\mathbb{R}$ equipotent to $\mathbb{R}^2$?

Answer : Yes

Hint : Since $\mathbb{R}$ is equipotent to $]0,1[$, you just have to prove $]0,1[$ is equipotent to $]0,1[^2$. Now you can use decimal expansion and the same kind of trickery you would use to show $\mathbb{N}$ and $\mathbb{N}^2$ are equipotent.

Personally I detest decimal expansions. I would be interested in other proofs than the standard one (which is in Joel's reply) that avoid this.

(By the way, to show that $|\mathbb{N}\times \mathbb{N}|=|\mathbb{N}|$ I prefer the argument that $(m,n)\mapsto 2^m3^n$ is injective by the fundamental theorem of arithmetic.)

For example, we could use that for an infinite field $K$, its algebraic closure has the same cardinality. (This follows because $|K^n|=|K|$ and then $|K[x]|=|\cup_{n\in\mathbb{N}}K^n|=|\mathbb{N}||K|=|K|$.) Hence $|\mathbb{R}|=|\mathbb{C}|=|\mathbb{R}^2|$.

Who knows a (edit: non-circular) argument?

A nice (general) argument is based on some elementary point-set topology, as described at wikipedia: any compact Hausdorff space with more than two point, and all of whose singletons are non-open, must be uncountable.