Above is my question. This is my progress so far (first part):
We know that the Borel $\sigma$-algebra $\mathcal B([0,1])$ is generated by (possibly infinite) unions of intervals of the form $(a,b)$, along with $[0,b)$ and $(a,0]$. Given any finite union, I can certain obtain the result: say the union is of $N$ parts, choose all the lower limits to be the same and set the upper limit (closed part) to be $\epsilon/N$ less (or more), we can have exactly $(a,0]$, and use $(\epsilon/2,b-\epsilon/2]$; this gives the measure of $A \Delta B$ to be $2\epsilon$, which is of course fine.
UPDATE: I can do the countable part fairly easily, actually - similarly to above, but instead of $b_n - \epsilon/N$, use $b_n - \epsilon/n^2$, which is fine since $\sum_{n \in \Bbb N} 1/n^2 (= \pi^2/6 ?) < \infty$, and scaled by $\epsilon$.
UPDATE 2: Since $\mathcal B$ is generated as above, we only need to consider countable unions. This is the bit that I am unsure about - see below. This statement here (Update 2) is what I am unsure about - can someone confirm this either way for me? Thanks!
I have an intuition as how to do it for the uncountable union (but it might not work) ignoring the $[0,b)$ and $(a,0]$ terms as they don't matter: given the infinite union, since $[0,1]$ is bounded in $\Bbb R$, and $\mu([0,1]) = 1$, we may write $$ B = \bigcup_{i \in I}(a_i,b_i) = \bigcup_{i \in I_1}(a_i,b_i) \cup \bigcup_{i \in I_2}(a_i,b_i), $$ where $I_1$ and $I_2$ are chosen so that $I_1$ is finite and $I_2$ is the "small tail" in the sense that $b_i - a_i$ is small for $i \in I_2$, and we can bound it, say by $1/i^2$ so that when we sum over it we get a finite number, multiplied by $\epsilon$, say.
Unfortunately, I can't get the details out, so I am hoping for some assistance! Thanks!
PS: Sorry for the awful title - if someone else can think of a better one, then please do let me know!
PPS: I have checked other questions for duplicates - sorry if this is a duplicate!