As was pointed out in the comments, your argument for (i) is perfectly fine.
Concerning (ii), yes, that's true, you could apply that result. On the other hand, the direct proof is straightforward:
We want to prove that
$$\tag{#}
\sup{
\left\{
\sum_{k=1}^n |f(a_k) - f(a_{k-1})|
\,:\,
0\leq a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq a_n \leq 1
\right\}
} \lt \infty
$$
where the supremum is taken over all finite increasing sequences in $[0,1]$. Finiteness follows from the Lipschitz condition $|f(x) - f(y)| \leq M\cdot|x-y|$ because for a finite increasing sequence $0\leq a_0 \leq a_1 \leq \cdots \leq a_{n-1} \leq a_n \leq 1$ we have
$$
\sum_{k=1}^n |f(a_k) - f(a_{k-1})| \leq \sum_{k=1}^n\phantom{|} M \cdot |a_k - a_{k-1}| \leq M
$$
because $|a_k-a_{k-1}| = a_{k}-a_{k-1}$ and $\sum_{k=1}^n (a_k-a_{k-1}) = a_n -a_0 \leq 1 - 0 = 1$. This proves that the supremum in $(\#)$ is in fact bounded by $M \lt \infty$.
This seems a bit easier than making the detour via absolute continuity (you'll notice that I hardly did anything else than what you did in your proof for (i), and that's why I gave the fully detailed argument).
Compare this with the proof of the fact that absolute continuity of a function $f$ implies that it is of bounded variation (which isn't much more difficult, but somewhat more fiddly, I think).
Now for (iii) I don't think that it is intended that you do this by hand, as a direct and detailed proof of this involves a considerable amount of work. So what you do is almost certainly the intention of this exercise.
Be a little careful, though: you say
I also know that if $f$ absolutely continuous $\implies$ $f$ is f bounded variation $\implies f'(x)$ exists a.e. and $$ f(x)=f(0)+\int_0^1 f'(x)\text{d}x. $$ Thus, $f(x)=\int_0^1 f'(x)\text{d}x~~~\text{if}~~f(0)=0.$
The second implication you write here is not true: bounded variation alone doesn't give that $f(x) = f(0) + \int_{0}^x f'(t)\,dt$ — note that there are two typos here as well:
- the upper bound of the integral should be $x$ instead of $1$ and
- your equality $f(x) = f(0) + \int_{0}^1 f'(x)\,dx$ involves $x$ with two different meanings: on the left hand side you have some point $x \in [0,1]$ and on the right hand side you have $x$ as a “dummy variable” for integration.
To see that the implication that bounded variation of $f$ doesn't imply that $f(x) = f(0) + \int_{0}^x f'(t)\,dt$, just notice that the Cantor function $h$ is monotone, hence of bounded variation, but $\int_{0}^x h(t)\,dt = 0$ for all $x \in [0,1]$.
However, I think that you meant to say that $f$ absolutely continuous implies $f(x) = f(0) + \int_{0}^x f(t)\,dt$ without interrupting by "$f$ has bounded variation", and then the rest of the argument here is fine too.
To get other (trivial examples) for your last question you can of course take any continuously differentiable (or even only absolutely continuous) function $g$ and add $h$ to it. Genuine examples not involving the Cantor function or simple variations of it are a bit cumbersome to come up with, so I think you solved that exercise satisfactorily, too.
When looking up something else, I recently found this nice and careful write-up by Noella Grady, which is about the present and many related matters.
Best Answer
Let $h = \mathbb 1_{(0,1]}$. For $x\in \mathbb R$ defines $$ g(x) = \sum_{n=1}^\infty 2^n h(x \cdot 2^n - 1) \frac{(-1)^{n+1}}{n}$$ and $$ f(x) = \int_0^x g(t) dt. $$
Then, $g$ is improper Riemann integrable and thus $f$ is continuous.
Further, on every compact interval without $0$, $g$ is bounded and thus $f$ is absolutely continuous (in fact Lipschitz).
But, on any neighborhood of $0$, $g$ is not Lebesgue integrable, thus $f$ is not absolutely continuous.
For the part that $f$ is additionally of bounded variation:
Notice that $f$ is AC if and only if $f$ is continuous, is BV, and maps measure zero set to measure zero set. Let $N\subset [0,1]$ have measure zero. Then, $N \cap [1/n, 1]$ has measure zero too and $$ |f(N)| = \lim_{n\to \infty} |f(N \cap [1/n, 1])| = 0. $$ That is, $f$ is AC on $[0,1]$.