Existence of everywhere continuous, nowhere differentiable function with finite multiplicity.

Does there exists a function $f:[0,1]\to[0,1]$ such that:
$(1)$$f$ is continuous everywhere.
$(2)$$f$ is differentiable nowhere.
$(3)$ For every $c \in range(f)$, there is only a finite number of points $x_1,...,x_k$ in $[0,1]$ such that, $f(x_i) = c$ for every $i\in \{ 0,..,k \}$.

I saw a related post about A-continuous-nowhere-differentiable-but-invertible-function. Although I am not sure if we can use that result to prove the non-existence of the above function because I don't know any measure theory. I would be grateful if someone showed any old post that dealt with this condition.