sphere-filling curve Let $S^2$ denote the $2$-dim sphere in $\mathbb R^3$. I am interested in finding a space-filling curve, i.e. a map $\varphi: [0,1]\to S^2$ that is continuous and onto.
We know that there is such a space-filling curve onto $[0,1]^2$ from Peano's and Hilbert's results.
Know my idea was to consider a unit cube in $\mathbb R^3$. Then I want to take a path that traverses enough edges of the cube (or just take a Hamiltonian path). From an edge I want to fill its face with Peano's curve and then go back to the edge after filling. This construction yields a cube-filling curve. Then I blow the cube up to the sphere and I am done.
However, this seems to simple to me. Does that work or do I miss something?
 A: Your construction seems doable. Here's an alternate rough sketch of a construction (I hope it's ok if I don't produce the details): 
Choose space filling curves $f_{m, n} : [0, 1] \to [n, n+1] \times [m, m + 1]$ for each integer $m, n$ so that they glue to a continuous surjection $f : \Bbb R \to \Bbb R^2$ in the sense that for each unit interval $I_k = [k, k +1] \subset \Bbb R$ with $k$ an integer, $f|_{I_k}$ is $f_{m, n}$ for some choice of $m, n$ (depending on $k$). That is, glue a bunch of space filling curves on each $[n, n + 1] \times [m, m + 1]$ systematically so that the endpoints agree. Continuity of $f$ is guaranteed by gluing lemma since endpoints of $f_{m, n}$ agree. 
Note that the above construction can be done because you can always choose a space filling curve $f : I \to I^2$ so that the endpoint $f(0)$ and $f(1)$ are whatever you want them to be (you can cook up a Peano or Hilbert like construction with endpoints given). 
Staring a bit carefully tells you that you can actually choose the endpoints so that $f$ is proper (that is, glue $f_{m, n}$ in a way so that $\{f(k)\}$ heads off to infinity as $k \to +\infty$ or $k \to -\infty$). Once that is done, proper maps can be extended to one-point compactifications, so $f$ extends to a space filling curve $\tilde{f} : S^1\to S^2$. Now compose with the map $g : I \to S^1$, $g(x) = e^{2\pi ix}$ and you're done. 
A: I would approach this by first trying to fill a 3D simplex with a space filling triangle like this: https://www.youtube.com/watch?v=pw_50szQfA0. Then subdivide the simplex to approximate a sphere.
We want to construct a function $f: [0,1) \mapsto S^2$ such that $f$ is bijective and preserves locality like Peano's and Hilbert's curves.
Let $v_k$ denote the vertices of a tetrahedron inscribed in a sphere of radius 1. That is $||v_k|| = 1, \forall k \in \{0,1,2,3\}$. 
Define $N(v) = \cfrac{v}{||v||}$
Let's first define the inverse map:
Given a vector $v$
First, find out the closest 3 points of the tetrahedron in the sphere. Say they are $v_a, v_b, v_c$. The first digit of the answer will be the only one in $\{0,1,2,3\} - \{a,b,c\}$
Now we subdivide this section, adding the points $N(v_a +v_b), N(v_b+v_c), N(v_c + v_a)$ to form 4 new triangles in the sphere:
(0) $N(v_c + v_a), v_a , N(v_a + v_b)$; 
(1) $N(v_a + v_b), N(v_b + v_c), N(v_c + v_a)$; 
(2) $N(v_a + v_b), v_b, N(v_b + v_c)$; 
(3) $N(v_b + v_c), v_c , N(v_c + v_a)$;
Then you find the 3 closest of these in the sphere and you know in what triangle you are, the next digit is given by the name of the triangle in brackets. Now rename the vertices of the inner triangle and repeat the process.
Define $f$ by:
Given $x \in [0,1)$
Let $x_k = \cfrac{\lfloor x\cdot 4^{k} \rfloor}{4}$ denote the k'th digit of x in base 4 after the radix. 
Firstly, with $x_1$ we already know that we are in the only triangle that doesn't contain $v_{x_1}$, now we get the other vertices $a,b,c$ of the triangle and subdivide it like in the definition of the inverse, the next digit determines in which triangle we are, rename the vertices of the newly find triangle in the same order provided and continue with the next digit
