Is there a surjective continuous $f : [0,1] \rightarrow {[0,1]}^{2}$. I know that There is such a function like this but the domain is cantor set.

There is a continuous surjective $f: C \rightarrow [0,1]$ where $C$ is the Cantor set. So $f \times f$ maps $C \times C$ onto $[0,1] \times [0,1]$. $C \times C$ is homeomorphic to $C$, so composing with the homeomorphism we can assume we have $g: C \rightarrow [0,1] \times [0,1]$ (you were this far already). Now apply Tietze-Urysohn to extend $g$ to $[0,1]$. Done. Note that we can pick $[0,1]^n$ or $[0,1]^\mathbb{N}$ as well, if we liked.
In fact (deeper theorem): every Peano continuum (i.e. metric compact connected and locally connected space) is the continuous image of $[0,1]$, and $[0,1]^2$ is a Peano continuum.
