The measure of the image of a set of measure zero

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 $f(E)\subseteq \cup f(w_k,z_k)$.