Your argument is basically correct. Notice that you use monotone (and increasing) so that $f(x,y)=(f(x),f(y))$ is true for all $x<y$ (a minor technicality is that it could happen that $f(x,y)$ is a closed or half-open interval unless $f$ is strictly increasing). Other than that, there are some places that could be clearer: "Let $(x_k,y_k)\subset [x_i,y_i]$ cover $E$." What exactly do you mean? (What is the relationship between $k$ and $i$? You mean to take a sequence of intervals, not just one, right? What is $[x_i,y_i]$ here?) In other words, the ideas are correct but could use some cleaning up and clarification, perhaps along the lines of:
Given $\varepsilon>0$, let $\delta>0$ be such that ....
Because $E$ has measure $0$, there exists a countable collection of disjoint intervals $\{(x_k,y_k)\}$ such that $E\subseteq \cup (x_k,y_k)$ and $\sum(y_k-x_k)<\delta$.
[Perhaps consider here the technicality that some of the $(x_k,y_k)$ may not be in $[0,1]$, which can be handled by intersecting everything with $[0,1]$, possibly leaving you with one or two half-open intervals on the end, but causing no problems.]
[Insert rest of your proof.]
Monotone is not necessary; the result would be true if $f$ were only assumed absolutely continuous. To see this, suppose the same setup with $E\subseteq \cup (x_k,y_k)$ and $\sum(y_k-x_k)<\delta$. For each $k$, because $f$ is continuous on $[x_k,y_k]$, there exists $w_k<z_k$ in $[x_k,y_k]$ such that $\{f(w_k),f(z_k)\}=\{\min\limits_{x\in[x_k,y_k]}f(x),\max\limits_{x\in[x_k,y_k]}f(x)\}$. Then you have $\sum(z_k-w_k)\leq\sum(y_k-x_k)<\delta$, so $\sum \mu(f(w_k,z_k))<\varepsilon$, and $f(E)\subseteq \cup f(w_k,z_k)$.
I don't think Zorich is super clear on this point. The closest he comes to an explanation is this:
The order of summation $\sum_{n,k} |I^n_k|$ on the indices $n$ and $k$ is of no importance, since the series converges to the same sum for any order of summation if it converges in even one ordering. [Zorich is referring to Proposition 4, page 270.] Such is the case here, since any partial terms of the sum are bounded above by $\varepsilon$.
I would explain this this way. Let $a_n = \sum_{k} |I^n_k| < \frac{\varepsilon}{2^n}$. Now let $A = \sum_{j=1}^J |I^{n_j}_{k_j}|$ be any partial sum of the series $\sum_{n,k} |I^n_k|$ arranged in an arbitrary order. If we let $N = \max_j n_j$, then $A \leq \sum_{n = 1}^N a_n \leq \sum_{n = 1}^{+\infty} a_n$.
Since any partial sum of the series $\sum_{n,k} |I^n_k|$ is $\leq \sum_{n = 1}^{+\infty} a_n$, we have
$$
\sum_{n,k} |I^n_k| \leq \sum_{n = 1}^{+\infty} a_n < \varepsilon.
$$
It's true there are theorems about double series that would make the original claim obvious, but you can get by without them.
Best Answer
Hint (for ii): Show that if $E$ is Jordan null, then the (topological) closure of $E$ is Jordan null. From this you can show that each Jordan null set is nowhere dense. Finally, show that $A$ cannot be written as a countable union of nowhere dense sets. (In fact, $A$ is easily large enough for this, since the complement of $A$ is a first category set.)
Additional Hints: Assume $E$ is Jordan null. Let $\varepsilon > 0$ be given. Cover $E$ with open intervals $I_{1}, \; I_{2}, \; ..., \; I_{n}$, where the sum of the lengths of these open intervals is less than $\frac{\varepsilon }{2}$. Take the closure of both sides of $E\subseteq \bigcup\nolimits_{n=1}^{N}I_{n}$, using the fact that the closure operator is monotone with respect to set inclusion and distributes over a finite union. This gives
$$\overline{E}\;\;\subseteq \;\;\overline{\bigcup\nolimits_{n=1}^{N}I_{n}} \;\; = \;\; \bigcup\nolimits_{n=1}^{N}\overline{I_{n}} \;\;= \;\; \left( \bigcup\nolimits_{n=1}^{N}I_{n}\right)\; \cup \;\mathcal{E},$$
where $\mathcal{E}$ is the finite set of endpoints of the intervals $I_{1}, \; I_{2}, \; ..., \; I_{n}$. Now note that $\mathcal{E}$ can be covered by a finite collection of open intervals whose lengths add to less than $\frac{\varepsilon }{2}$.