Reparameterisation of Curve as a Regular Curve (Topology) There is a result that a curve or topological path can be reparameterized as a regular curve contained in the paper "Reparametrizations of continuous paths - Ulrich Fahrenberg and Martin Raussen" https://arxiv.org/pdf/0706.3560.pdf 
For me the concepts are too advanced. Is there a simpler proof for this ? 
The definitions that follow are taken from the paper (more or less).
A path is a continuous mapping $p$ from the closed unit interval $I = [0, 1]$ to a topological space $X$.
Excluding the case that the image of a path is a single point in X, then a path is regular if there is no closed interval $[a, < b] \subset I$ on which $p$ is constant.
A reparametrization $\phi$ is a non-decreasing surjective continuous map $\phi: I \to I$ with $\phi(0) = 0; \phi(1) = 1$.
Theorem: for any path $p: I \to X$ there is a regular path $q: I \to X$ and a reparametrization $\phi$ such that $p = q \circ \phi$.

It looks straightforward to prove this in the case that $p$ has a finite number of stop intervals (on which $p$ is constant) by cutting them out one by one and composing the corresponding $\phi$ functions. Clearly the number of stop intervals is countable, but how to deal with a countably infinite number of them ?
 A: (*x) refer to footnotes added to the original answer...

You exclude the case that $p : I \to X$ is constant. Let $\mathcal C$ the set of all components (*a) of all preimages $p^{-1}(x)$ with $x  \in p(I)$ and $\mathcal S$ the subset of all $S \in \mathcal C$ having more than than one point. Since $S$ is a connected subset of $I$, it is an interval. Since we require that $S$ has more than one point, it has length $> 0$ and we call it a stop interval of $p$. It may be an open, half-open or closed interval.
In the sequel we assume that all stop intervals are closed intervals. This is automatically satisfied if $X$ is a $T_1$-space (*b). Then all $p^{-1}(x)$ are closed in $I$ so that also all of its components are closed. If there exists a non-closed stop interval, some arguments below are no longer valid.
The set $\mathcal S$ is countable (*c) (either finite, including empty, or infinite). Let $C = \bigcup_{S \in \mathcal S} S$. Let $\mathcal T$ denote the set of components of $I \setminus C$. Each element of $\mathcal T$ is an interval  - open, half-open or closed (which may be degenerate to point). (*d)
For each closed subinterval $J = [c,d] \subset I$ let
$$\mathcal S_J = \{ S \cap J \mid S \in \mathcal S, S \cap J \ne \emptyset \} .$$
This is again a countable set of closed intervals. It is possible that $S \cap J =  \{c\},\{d\}$ (degenerate intervals), but this does not matter. The number
$$\lvert \mathcal S_J \rvert = \sum_{A \in \mathcal S_J} \lvert A \rvert$$
is well defined. Here $\lvert A \rvert$ denotes the length $b - a$ of the interval $A = [a, b]$. Obviously we always have $\lvert \mathcal S_J \rvert \le \lvert J \rvert$.
Define
$$s : I \to I, s(t) = \lvert \mathcal S_{[0,t]} \rvert .(*e)$$
For $t \le t'$ we have $s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert$. To verify this, let $S = [a,b] \in \mathcal S$ such that $S \cap [0,t'] \ne \emptyset$. We have $S \cap [0,t'] = (S \cap [0,t]) \cup (S \cap [t,t'])$. If one these two intersections is empty, then the summand $\lvert S \cap [0,t'] \rvert$ of $\lvert \mathcal S_{[0,t']} \rvert$ occurs in exactly one of the sums $\lvert \mathcal S_{[0,t]} \rvert$, $\lvert \mathcal S_{[0,t]} \rvert$. If both intersections are nonempty, then $\lvert S \cap [0,t'] \rvert = \lvert S \cap [0,t] \rvert + \lvert S \cap [t,t'] \rvert$.
We conclude that


*

*For $t \le t'$ we have $s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert \le s(t) + (t' - t)$.

*$s$ is continuous because 1. implies $\lvert s(t') - s(t) \rvert \le \lvert t' - t \rvert$ for all $t, t'$.

*The restriction of $s$ to any $S = [a,b] \in \mathcal S$ has the form $s(t) = s(a) + \lvert \mathcal S_{[a,t]} \rvert = s(a) + \lvert [a,t]\rvert  = s(a) + (t-a)$.

*The restriction of $s$ to any $T \in \mathcal T$ is constant (with some value $c_T$). To see this, let $t,t' \in T$ with $t \le t'$. Then $s(t') = s(t) + \lvert \mathcal S_{[t,t']} \rvert = s(t)$ because $[t,t'] \subset T \subset I \setminus C$ does not intersect any $S \in  \mathcal S$.
Define $r(t) = t - s(t)$. This is a continuous function such that $r(t) \ge 0$ and $r(0) = 0$. The function $r$ is non-decreasing because for $t \le t'$ we have $r(t') - r(t) = t' - s(t') - (t - s(t)) = (t' - t) - (s(t') - s(t))  \ge 0$ (see 1.). Since $r(1) = 1 - s(1)$, we regard $r$ as continuous non-decreasing surjection $r : I \to [0,1-s(1)]$.
Moreover, for $t\in S = [a,b] \in \mathcal S$ we have $r(t) = t - s(a) - (t-a) = a - s(a)$ and for $t \in T \in \mathcal T$ we have $r(t) = t - c_T$.  This means that the stop intervals of $r$ are the same as those of $p$. Hence $p = p' \circ r$ with a unique function $p' : [0,1-s(1)] \to X$. Because $I$ is compact, $r$ is a quotient map and $p'$ is continuous (*f, *g, *h). Stretching $[0,1-s(1)]$ to $I$ yields the desired result. Note that $s(1) )= 1$ is impossible because in that case $p$ would be constant.
Let us finally see where the argument breaks down if there exists a non-closed stop interval $S$. Then $S \subsetneqq \overline{S} = [a, b]$ and 3. holds on $\overline{S}$. Hence $r$ is constant on $\overline{S}$ (this follows also from continuity). Thus $r$ and $p$ do not have the same stop intervals. As an example consider any non-constant map $f : I \to \{0,1\}$, where $\{0,1\}$ has the trivial topology, such that $p(x) =  0$ for $x < 1/2$ and $p(x) = 1$ for $x \ge 1/2$. Then $p$ has stop intervals $[0,1/2)$ and $[1/2,1]$, but $r$ has stop interval $I$.
Remark:
One could also use the definition
$$s(t) = \int_0^t \chi_C(x)dx$$
where $\chi_C$ is the characteristic function of the subset $C \subset I$ (i.e. $\chi_C(x) = 1$ for $x \in C$, $\chi_C(x) = 0$ for $x \notin C$). However, a proof that $\chi_C$ is integrable is needed..

Clarifications and references ...
(*a) implicit in the definition of "components" is that they are connected.
(*b) A space is $T_1$ if and only iff every singleton point set is closed.
(*c) $\mathcal S $ must be countable in order for the sum of positive lengths of $S \in \mathcal S$ to be finite.
(*d) Although each $S$ is closed the (countably) infinite union need not be closed - hence the possible different type of interval in $\mathcal T$.
(*e) $s(t) $ is then the total "stopped length" up to the point $t$.
(*f) Munkres - Topology, p.135 definition of quotient map: a continuous surjective closed map is a quotient map.
(*g) https://math.stackexchange.com/q/548598 -  continuous map from compact space to Hausdorff is closed.
(*h) Munkres - Topology, p.142 Theorem 22.2 - commutivity diagram: existence and continuity of $p'$
