Setting: Let $F':[0,1] \rightarrow [0,1]$ be the Cantor Function.
Goal: Show that there exists a well-defined, surjective, continuous function from $[0,1]$ to $[0,1]^2$ (i.e., a space-filling curve).
EDIT: It turns out my function is non-sensical in that the domain isn't even $[0,1]$! I'm leaving it here to show an attempt was made at answering the question, but I've yet to come up with a suitable candidate function from $[0,1]$ to $[0,1]^2$.
Let $G: [0,1] \rightarrow [0,1] \times [0,1]$ s.t.
$$ G(x,y) = (F'(x),F'(y)) $$
I claim that $G$ is a well-defined, surjective, continuous map.
First $G$ is clearly well-defined (there are no issues with representatives or equivalence classes).
To see that $G$ is surjective, let $(a,b) \in [0,1]^2$.
Consider $F'$ is surjective (I've already shown this on my own).
Hence we have that $\exists x,y \in [0,1]$ s.t. $F'(x) = a$ and $F'(y) = b$.
Then consider that $G(x,y) = (F'(x), F'(y)) = (a,b)$ so that $G$ is surjective as desired.
Question: Why is $G$ continuous? I've shown elsewhere that $C \cong [0,1]$, and I've been given the hint that $C \cong C \times C$ would be a useful fact to prove. Why would this fact be useful?