The set of zero measure from Zorich's book This is actually from Zurich's book (chapter on Integration).

Definition. A set $E\subset \mathbb{R}$ has measure zero or is of measure zero (in the sense of Lebesgue) if for every number
$\epsilon>0$ there exists a covering of the set $E$ by an at most
countable system $\{I_k\}$ of intervals, the sum of whose lengths
$\sum \limits_{k=1}|I_k|$ is at most $\epsilon$.
Lemma. The union of a finite or countable number of sets of measure zero is a set of measure zero.

Proof: Suppose that $E=\cup_{n=1}^{\infty}E_n$ and each $E_n$ has measure zero. Suppose $\epsilon>0$ be given. For each $n\in \mathbb{N}$ by definition we can find an at most  countable system $\{I_{n,k}\}_{k\geq 1}$ of intervals such that $\sum \limits_{k=1}^{\infty}|I_{n,k}|<\dfrac{\epsilon}{2^n}$.
Then we see that $\{I_{n,k}: n,k\geq 1\}$ be a countable system of intervals covering $E$.
How formally to show that total length of these $I_{n,k}$ is less than $\epsilon$?
Because if we want to compute its total length then we need to deal with the double sum $\sum \limits_{n,k\geq 1}$ or $\sum \limits_{n\geq 1} \sum \limits_{k\geq 1}
$ (btw I don't know the difference between them).
I would be very thankful if someone can show the formal proof of it please!
 A: 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.
A: The point here is that you don't actually need to compute it.  What you do is for your first set,  $E_1$,  choose a covering so that the total measure is less than $\frac \epsilon 2$.  The second set you choose one whose total is less than $\frac \epsilon 4$,   then $n'th$ one a cover whose total length is less than $\frac \epsilon {2^n}$.   Now,   taking the union and all you have is the simple geometric series $\frac \epsilon {2^n}$,  which adds up to $\epsilon$
Edited to add complicated notation:
For Set $E_1$  choose a covering $E_{1,k}$  such that $\sum_{k\geq 1}E_{1,k}<\frac \epsilon 2$.   Similarly, for each set $E_n$ choose a covering so that $\sum_{k\geq 1}E_{n,k}<\frac \epsilon {2^n}$
Now,  $\bigcup_{n\geq 1}E_n $is covered by $\bigcup_{n\geq 1}\bigcup_{k\geq 1}E_{n,k}$,  and taking the sum of the lengths,  we have the union is less than
$$\sum_{n\geq 1}\sum_{k\geq 1}E_{n,k}=\sum_{n\geq 1}\left(\sum_{k\geq 1}E_{n,k}\right)=\sum_{n\geq 1}\frac \epsilon {2^n}=\epsilon $$
