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You might want to look into space filling curves, which were first described by Peano and Hilbert in the late 1800's. These are continuous surjections from $[0,1]$ onto $[0,1]^2$ (and higher powers) but they are not bijections. However, they are visualizable to a certain extent. A quick Google search gave a lot of hits, in particular this one at Cut The Knot which is not necessarily the best but it has an illustrative java applet.

As for the existence of a bijection, you can derive it from the fact that $\aleph_0\cdot2 = \aleph_0$ and the usual exponent rules: $$(2^{\aleph_0})^2 = 2^{\aleph_0\cdot2} = 2^{\aleph_0}$$ It is also easy to write an explicit bijection between Cantor space $\{0,1\}^{\mathbb{N}}$ (the space of infinite binary sequences) and its square by splitting the even and odd coordinates. This, together with a bijection between $\mathbb{R}$ and $\{0,1\}^{\mathbb{N}}$, gives what you want. Note that it is this last bijection which is harder to visualize. The reason is that $\mathbb{R}$ is connected while $\{0,1\}^{\mathbb{N}}$ is totally disconnected (with the product topology).

You might want to look into space filling curves, which were first described by Peano and Hilbert in the late 1800's. These are continuous surjections from $[0,1]$ onto $[0,1]^2$ (and higher powers) but they are not bijections. However, they are visualizable to a certain extent. A quick Google search gave a lot of hits, in particular this one at Cut The Knot which is not necessarily the best but it has an illustrative java applet.
As for the existence of a bijection, you can derive it from the fact that $\aleph_0\cdot2 = \aleph_0$ and the usual exponent rules: $$(2^{\aleph_0})^2 = 2^{\aleph_0\cdot2} = 2^{\aleph_0}$$ It is also easy to write an explicit bijection between Cantor space $\{0,1\}^{\mathbb{N}}$ (the space of infinite binary sequences) and its square by splitting the even and odd coordinates. This, together with a bijection between $\mathbb{R}$ and $\{0,1\}^{\mathbb{N}}$, gives what you want. Note that it is this last bijection which is harder to visualize. The reason is that $\mathbb{R}$ is connected while $\{0,1\}^{\mathbb{N}}$ is totally disconnected (with the product topology).
You might want to look into space filling curves, which were first described by Peano and Hilbert in the late 1800's. These are continuous surjections from $[0,1]$ onto $[0,1]^2$ (and higher powers) but they are not bijections. However, they are visualizable to a certain extent. A quick Google search gave a lot of hits, in particular this one at Cut The Knot which is not necessarily the best but it has an illustrative java applet.
As for the existence of a bijection, you can derive it from the fact that $\aleph_0\cdot2 = \aleph_0$ and the usual exponent rules: $$(2^{\aleph_0})^2 = 2^{\aleph_0\cdot2} = 2^{\aleph_0}$$ It is also easy to write an explicit bijection between Cantor space $\{0,1\}^{\mathbb{N}}$ (the space of infinite binary sequences) and its square by splitting the even and odd coordinates. This, together with a bijection between $\mathbb{R}$ and $\{0,1\}^{\mathbb{N}}$, gives what you want.