Do there exist space-filling curves that fill the whole plane? If so, can they be visualized?

Update: Here is a neater construction!

As Daniel Fischer has pointed out this cannot be done with $[0,1]$ as parameter interval. Therefore we shall map the half-open interval $[0,1[\ $ continuously onto ${\mathbb R}^2$. Use the redraw rule shown in the above figure to create a Peano curve $$\gamma_*: \quad\bigl[0,1]\to R:=[0,\sqrt{3}]\times[0,1]\ .$$ (Sketch of proof: Let $t\mapsto\phi_0(t)\in R$ be the parametrization of the diagonal in the lefthand figure above. The redraw rule, applied recursively in nested triadic subintervals of $[0,1]$, produces a sequence $(\phi_n)_{n\geq0}$ of piecewise linear maps $\phi_n:\>[0,1]\to R$. It is easy to check that $$|\phi_{n+1}(t)-\phi_n(t)|\leq 3^{-n/2}\qquad(0\leq t\leq1)\ ,$$ and this implies that the $\phi_n$ converge uniformly to a continuous map $\gamma_*:\>[0,1]\to R$. Since $\gamma_*(I)$ is closed and dense in $R$ it has to be all of $R$.)

Stretch this $\gamma_*$ horizontally by a factor $\sqrt{2\over3}$, so that we now have a Peano curve $$\gamma_0:\quad\bigl[0,1]\to[0,\sqrt{2}]\times[0,1]\ ,$$ beginning at $(0,0)$ and ending at $(\sqrt{2},1)$. This $\gamma_0$ is now concatenated with a copy of itself, then with ever larger copies scaled by factors $2^{n/2}$, in a spiraling way, see the following figure which shows only the "diagonal" of each copy:

For the concatenation we allot the time interval $\bigl[0,{1\over2}\bigr]$ for $\gamma_0$, then the subsequent time intervals $$\bigl[1-2^{-n},1-2^{-(n+1)}\bigr]\quad(n\geq1)$$ for the subsequent rectangle filling curves.