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"?

2$\begingroup$ If I'm not wrong the Peano square fillingcurve covers any point at most $4$ times; this follows easily from the analytic description of it $\endgroup$ – Pietro Majer Aug 6 '18 at 7:27

2$\begingroup$ You may look at ac.elscdn.com/S0166864197002708/… They construct a map from $S^1$ to $S^2$, but you may restrict to a subinterval. More generally, the theory of CannonThurston maps constructs Peano curves, where preimages of points typically seem to have cardinality $1$ or $2$. $\endgroup$ – ThiKu Aug 6 '18 at 7:32
Recall the definition of the Peano squarefilling 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. $$b_n:={\bf k}^{a_2+a_4+\dots a_{2n2}}a_{2n1}$$ $$c_n:={\bf k}^{a_1+a_3+\dots a_{2n1}}a_{2n}$$ where ${\bf k}$ is the involutory bijection of the set $\{0,1,2\}$ into itself given by $i\mapsto2i$ (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 $$a_{2n1}:={\bf k}^{c_1+c_2+\dots +c_{n1}}b_{n}$$ $$a_{2n}:={\bf k}^{b_1+b_2+\dots +b_{n}}c_{n}.$$
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 iff $y\in (T^c \times T)\cup (T\times T^c) $, four elements iff $y\in T\times T$. This is also mentioned in Peano original short paper.

2$\begingroup$ One can modify the construction to make $f^{1}\{y\}\le 3$; in general a space filling curve in $d$dimensional space (for the covering dimension) has points of multiplicity at least $d+1$ and this bound is optimal. $\endgroup$ – Anton Petrunin Aug 6 '18 at 16:26

$\begingroup$ @PietroMajer Thank you very much for your very great answer. $\endgroup$ – Ali Taghavi Aug 7 '18 at 14:10