Is there a continuous surjective function $f:[0,1] \to [0,1]^2$ such that
every level set $f^{-1}(y)$ is a finite set? If the answer is no, what about if we replace the finiteness of level sets by "countable level sets"?



Space filling curve whose all level sets are finite (countable)
Recall the definition of the Peano square-filling curve $f:[0,1]\to[0,1]^2$, which is given in terms infinite ternary strings. If $a\in [0,1]$ has a base $3$ representation of the form $0,a\_1a\_2a\_3\dots$, the point $f(a):=(b,c)$ has base $3$ digits resp. $$b\_n:={\bf k}^{a\_2+a\_4+\dots a\_{2n-2}}a\_{2n-1}$$
$$c\_n:={\bf k}^{a\_1+a\_3+\dots a\_{2n-1}}a\_{2n}$$
where ${\bf k}$ is the involutory bijection of the set $\{0,1,2\}$ into itself given by $i\mapsto2-i$ (and exponents denote iterated composition, which in this case only depends on the parity of the exponent, since ${\bf k}^2={\bf id}$). As Peano observes, this gives a bijection (actually a homeomorphism wrto product topologies of discrete spaces) $\{0,1,2\}^{\mathbb{N}} \to \{0,1,2\}^{\mathbb{N}}\times \{0,1,2\}^{\mathbb{N}}$. In fact
$$a\_{2n-1}:={\bf k}^{c\_1+c\_2+\dots +c\_{n-1}}b\_{n}$$
$$a\_{2n}:={\bf k}^{b\_1+b\_2+\dots +b\_{n}}c\_{n}.$$


The point of the whole construction is that, thanks to the effect of the map ${\bf k}$, the above bijection on ternary strings is compatible with the quotient map $\operatorname{val}:\{0,1,2\}^{\mathbb{N}} \to[0,1]$, that takes a ternary string $(a\_1,a\_2,\dots)$ to its value as a ternary expansion of a real number, $\sum\_{n=1}^\infty 3^{-n}a\_n$. The latter map $\operatorname{val}$ is surjective but of course not injective, due to points in $[0,1]$ with double representations of "ternary rationals", that is points in the set $T:=\{ m/3^r: r\in\mathbb{N},\ 0<m<3^r\}$. So passing to the quotient produces a (continuous, surjective, not injective) map $f:[0,1]\to[0,1]^2$. 


Now, going back to the question: it is immediate by the construction, that $f^{-1}(y)$ is a single point iff $y\in T^c\times T^c$; it has two elements if $y\in (T^c \times T)\cup (T\times T^c) $, and it has either two or four elements if $y\in T\times T$. 
This is also mentioned in Peano original short paper. 


