The result that $f$ is strictly increasing holds for any non-decreasing and continuous function $\phi$ satisfying $\phi(0)=0$ and $\phi(1)=1$, not just for the Cantor-Lebesgue function.
Here is an example that if $\phi$ is not continuous, then $f$ need not be strictly increasing. Let $\phi(t)=0$ for $t<\frac12\,,$ let $\phi(t)=t$ for $t>\frac12$
(and let $\phi(\frac12)$ be any value between $0$ and $\frac12$ inclusive). Then $f(0)=\frac12=f(\frac12)$, and $f$ is constant $f\equiv\frac12$ on the interval $[0,\frac12]$, so $f$ is not strictly increasing.
Now assume that $\phi$ is any non-decreasing and continuous function on $[0,1]$
with $\phi(0)=0$ and $\phi(1)=1$.
We will show that $\phi(f(x))=x$ for every $x\in[0,1]$.
We have $\phi(f(x))\le x$ by continuity of $\phi$.
Indeed, if $\phi(f(x))>x$ then there is some $\varepsilon>0$ such that
$\phi(f(x)-\varepsilon)>x$.
(Note that $f(x)>0$ since $\phi(f(x))>x\ge0=\phi(0)$.)
If we let $s=f(x)-\varepsilon$ then
$\phi(s)>x$, hence $f(x)\le s$ by the definition of $f(x)$,
obtaining a contradiction $f(x)\le f(x)-\varepsilon$.
Next we show that $\phi(f(x))\ge x$. If $\phi(f(x))<x$ then
by continuity of $\phi$ there is some $\varepsilon>0$ such that
$\phi(f(x)+t)<x$ for all $t\in[f(x),f(x)+\varepsilon)$,
resulting in the contradiction $f(x)\ge f(x)+\varepsilon$
(again using the definition of $f(x)$).
(Note that $f(x)<1$ since $\phi(f(x))<x\le1=\phi(1)$.)
It follows that if $0\le a<b\le1$ then $0\le f(a)<f(b)\le1$.
Indeed $\phi(f(a))=a<b=\phi(f(b))$, hence $f(a)\neq f(b)$, and it was already proved by the OP that $f(a)\le f(b)$.
The rest of this answer deals specifically with the
Cantor-Lebesgue function and treats the remaining questions.
$f$ is not continuous because of the horisontal line segments on the graph of $\phi$. For example, if $x<\frac12$ then $f(x)<\frac13$, and it is easily seen that
$\lim\limits_{x\to\frac12^-}f(x)=\frac13\neq f(\frac12)=\frac23$. On the other hand, $f$ is continuous from the right (using that in the definition of $f(x)$ we work with the inequality $\phi(t)\le x$ rather than $\phi(t)<x$).
The range of $f$ is a subset of the Cantor (middle-third) set $C$. Indeed, if $(p,q)$ is any of the "middle-third" intervals removed in the construction of $C$, then the range of $f$ misses $[p,q)$. We have that $p<q$ and $(p,q)\times\{c\}$ is a subset of the graph $G$ of $\phi$ (and $p$ and $q$ are some fractions with denominators $3^n$, and $c$ is a fraction with denominator $2^n$). Then $\lim\limits_{x\to c^-}f(x)=p\neq q=f(c)=\lim\limits_{x\to c^+}f(x)$.
It follows that $f$ is not measurable. Indeed take any set $M\subset[0,1]$ that is not measurable. Then $f(M)\subset C$ hence it has measure $0$ (as a subset of the measure $0$ Cantor set $C$), in particular it is measurable. The function $f$ is $1-1$ (being strictly increasing), hence $M=f^{-1}(f(M))$ showing that $f$ is not measurable.
One could also easily sketch the graph of $f$, at least as easy as one could sketch the graph $G$ of the Cantor-Lebesgue function $\phi$. The function $f$ is "almost" an inverse function of $\phi$. We obtain the graph of $f$ by the following steps. (a) reflect the graph $G$ of $\phi$ about the diagonal $y=x$, (b) remove all but the very top point of each of the vertical intervals that we obtain this way (vertical intervals that result from the reflection about $y=x$ of the horizontal intervals on the graph $G$ of $\phi$).
Here is a picture (from Wikipedia) of the graph of the
Cantor-Lebesgue function $\phi$.
Following is a picture of the graph of the
Cantor-Lebesgue function reflected about the line $y=x$.
Finally, a picture of the graph of $f$.
Best Answer
For present purposes, the following suffices:
Lemma: Suppose $[c, d] \subset [a, b]$ are closed intervals, that $f$ is a continuous, non-decreasing function on $[a, b]$, and that $g$ is a continuous function on $[a, b]$ satisfying:
$g(c) = f(c)$ and $g(d) = f(d)$;
$g$ is non-decreasing on $[c, d]$.
Then $g$ is non-decreasing on $[a, b]$.
To check that $x \leq y$ implies $g(x) \leq g(y)$, split into three cases: $x \leq c$; $d \leq y$; and $c \leq x \leq y \leq d$.
Each function $f_{n+1}$ in your sequence is obtained from $f_{n}$ by two operations of this type (given by your second set of formulas, with $1$ replaced by $n$ and $2$ replaced by $n + 1$), so $f_{n+1}$ is non-decreasing if $f_{n}$ is.