Yes, every curve parametrized by $[0,1]$ has an irreducible parametrization, and your ideas for proving this are mostly correct.
You have a lot of good ideas mixed in with a few that won't quite fly, but luckily the correct ideas are enough. That is, we can still make your inductive property approach work.
The problems.
The problematic part of your outline is that you don't know that you can decompose into countably many disjoint closed intervals in the way you describe. That is, even when $n=1$ and you consider the Cantor function, this function is constant on a countable union of intervals, and what is left over is not a countable union of closed intervals - rather it is a Cantor set. For the same reason your property $P$ is not inductive, as the Cantor set arises as an intersection of nested finite unions of disjoint closed intervals.
The easy part.
Firstly, we should get the easy part of the proof out of the way. Let $t\in[0,1]$ be the minimum value (which exists by compactness) such that $\gamma([0,t])=C$, then let $s\in[0,t]$ be the maximum value such that $\gamma([s,t])=C$. Without loss of generality we will take $s=0$, $t=1$, as we may certainly restrict and reparametrize $\gamma$ to ensure this. Then $\gamma([a,b])\neq C$ for any proper closed interval $[a,b]\subset [0,1]$.
It can be verified easily throughout the rest of this proof that none of our subsequent reparametrizations change this fact, and so we leave it to the reader to keep us honest while we find a reparametrization $\gamma_0$ such that $\gamma_0([0,1]\backslash (a,b))\neq C$ for all proper subintervals such that $\gamma_0(a)=\gamma_0(b)$.
Defining $P$ correctly.
Now define $P(X)$ to be the property that $X\subseteq [0,1]$ is closed, satisfies $\gamma(X)=C$, and satisfies $\gamma(a)=\gamma(b)$ for each component $(a,b)$ of $[0,1]\backslash X$.
We claim $P$ is inductive. To see this, suppose $X_i\supseteq X_{i+1}$ is a decreasing sequence with this property. Then certainly the intersection $X=\cap_{i=1}^\infty X_i$ is closed and satisfies $\gamma(X)=C$ (the latter can be seen from the fact that for each $c\in C$ the sets $\gamma^{-1}(\{c\})\cap X_i$ are nonempty, and so give a nested sequence of nonempty compact sets).
It remains to show that $\gamma(a)=\gamma(b)$ for each complementary interval $(a,b)\subseteq [0,1]\backslash X$. To see this, we suppose $0<\epsilon<\frac{b-a}{2}$. Denote $U_i=[0,1]\backslash X_i$, so that $\bigcup_{i=1}^\infty U_i=[0,1]\backslash X$. The sets $U_i$ form an open cover of $[a+\epsilon,b-\epsilon]$, hence have a finite subcover, which by monotonicity can consist of a single $U_i$. Then if $(a',b')$ is the component of $[a+\epsilon,b-\epsilon]$ in $U_i$, then since $U_i\cap X=\emptyset$, we have $$[a+\epsilon,b-\epsilon]\subseteq (a',b') \subseteq (a,b),$$
so that $a'\in [a,a+\epsilon)$ and $b'\in (b-\epsilon,b]$. Since $\gamma(a')=\gamma(b')$, letting $\epsilon\to 0$ gives $a'\to a$ and $b'\to b$, so we have $\gamma(a)=\gamma(b)$ by continuity of $\gamma$.
Completing the proof.
Having established that $P$ is inductive, let $X$ have property $P$ irreducibly, as per Brouwer's Theorem.
Your idea now works perfectly. Define $\gamma'$ to equal $\gamma$ on $X$ and to equal the common value $\gamma(a)=\gamma(b)$ constantly on every complementary component $(a,b)\subseteq [0,1]\backslash X$. Observe that if $[a,b]\subset [0,1]$ is any proper subinterval with $\gamma'(a)=\gamma'(b)$, and $(a,b)\cap X\neq\emptyset$, then $$\gamma'([0,1]\backslash (a,b))=\gamma(X\backslash (a,b))\neq C,\tag{1}$$ since otherwise we could let $X'=X\backslash (a,b)$ and obtain a smaller set with property $P$.
Now let $\gamma_0$ be a light mapping such that $\gamma' = \gamma_0\circ m$ for some monotone increasing surjection $m\colon [0,1]\to [0,1]$.
To complete the proof we must show that $\gamma_0([0,1]\backslash (a,b))\neq C$ for any proper subinterval $[a,b]$ with $\gamma_0(a)=\gamma_0(b)$. To see this, observe first that if we let $(a',b')=m^{-1}((a,b))$, (note that the inverse image of an interval under a monotone map is another interval), then we must have $(a',b')\cap X\neq\emptyset$, as otherwise $\gamma'$ is constant on $(a',b')$, implying by lightness of $\gamma_0$ that $m$ is constant on $(a',b')$, so that $m((a',b'))\neq (a,b)$, contradicting surjectivity of $m$.
But then we have $\gamma'(a')=\gamma'(b')$ and so from (1) we obtain $$\gamma_0([0,1]\backslash (a,b))= \gamma_0(m([0,1]\backslash (a',b')))=
\gamma'([0,1]\backslash (a',b'))\neq C.$$
Best Answer
(*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 $p'$