Let us define a curve as a continuous function $\phi:[a,b] \to \Bbb R^n$ and let us say that $\phi$ is simple if and only if $$\forall t,s \in ]a,b[, t \neq s \implies \phi(t) \neq \phi(s)$$ that is, $\phi$ is injective in $]a,b[$.
Now, I would like to know wether given a non-simple curve is it possible or not to split it into at most countably many simples curves. That is, given $\phi:[a,b] \to \Bbb R^n$, is it possible to write it as the concatenation of a collection of simple curves $\{\phi_i: [a_i, b_i] \to \Bbb R^n\}_{i \in \Bbb N}$ such that $a_0 = a$, $b_n \to b$ and $a_{i+1}=b_i$?
As always, any comment or answer is much appreciated and let me know if I can explain myself clearer!
Here is a counter-example that uses a space-filling continuous curve.
It is an example of a continuously parametrized curve in the plane that crosses itself at times that are dense in the time interval $0<t<1$.
Lemma. Every real number $t\in (0,1)$ has one and only one binary expansion $t= .a_1 a_2 a_3 \ldots$ that contains an infinite number of zeros.
This expansion is constructed as follows: start with the disjoint bisection $[0,1)=[0,1/2) \cup [1/2,1)$ (which are labeled as 0 and 1 respectively). Then bisect each of these subintervals in a similar way $\ldots$.
This construction gives the binary expansion of $t$ that is cheapest in the sense that it uses the fewest possible number of $1$'s. Note that if $t$ also admits an expensive binary expansion that has only a finite number of zeros, say $t=.00 111111 \ldots$, then $t=1/4$ can also be written as $t=.010000 \ldots$ instead, and the latter is the sort of expansion that is generated by the construction in the lemma.
Construction of the continuous curve.
For each $t= .a_1 a_2 a_3 \ldots$ set
$x(t)= .a_1 a_3 a_5 \ldots$ and
$y(t)= a_2 a_4 a_6 \ldots$.
In effect we deal out the digits of $t$ in alternating fashion to create an "odd pile" and an "even pile".
An easy estimate shows that $x(t)$ and $y(t)$ are both continuous functions of $t$. (For example, to know $x(t)$ and $y(t)$ to an accuracy of 7 decimal places, it suffices to know $t$ to an accuracy of 14 decimal places.)
Claim. This is a space-filling curve. (The curve maps onto a two-dimensional unit square.)
Proof. Consider any ordered pair $ (x,y)$ in the square $(0,1]\times(0,1]$. Use the lemma to write $x$ in binary and likewise $y$. Then merge their two expansions to obtain an expansion for a number that we call $t$. Then verify easily that our curve maps $t$ onto $(x,y)$.
Now let's construct some examples of times $t$ at which the curve crosses itself. Note that it is possible that $x(t)$ happens to have only a finite number of zeros (if all but finitely many of the 0's in $t$ are in the even pile $y(t)$ instead). In this case $x(t) =.**01111\ldots$ has a second binary representation that is terminating $x(t)=.**10\ldots$. Now merge these expansions $(x,y)$ to construct a different value $t' \ne t$ that satisfies $x(t')= x(t)$ and also $y(t')= y(t)$
Conclusion: The curve crosses itself at every special time $t$ for which the binary expansion of $t$ has an even or odd pile that contains only finitely many zeros. Note that every $t\in (0,1)$ is "arbitrarily close" to such special values, since we can tweak the binary expansion of $t$ at say only past the 100th decimal place to make it special.
That is, the special times at which the curve crosses itself are dense. Thus you cannot find any open subinterval in time that is free of self-crossings.