Walter Rudin's proof: countable union of countable sets is countable 
The capture is from Rudin's Principles of Mathematical Analysis, and I've seen similar proof for this theorem but with a different technique. It uses a single arrow to throw all the elements, the arrow can wiggle and turn around, this kind of proof I could understand.
Q1: When I see Rudin's proof, I was confused about how those arrows have bijections with $\mathbb{N}$. It said arrange in sequence, it seems that the sequence's terms are increasing(e.g $2$nd term is a two-element tuple, $3$rd term is a three-element tuple, etc.) can a sequence have terms for different types?
 A: A sequence can be whatever you want - but in this case, that's making things unnecessarily complicated. Just flatten the sequence-of-sequences down into a single sequence by following the arrows and concatenating the elements you pass by together in order. That is, trade the semicolons in (17) for commas, and discard the repeated entries.
A: 
it seems that the sequence's terms are increasing(e.g $2$nd term is a
two-element tuple, $3$rd term is a three-element tuple, etc )

This is incorrect.
Rudin is referring to the following sequence, which is a bijection of $\mathbb{N}$:
$x_{1,1}$  is the first term, $x_{2,1}$ is the second term, $x_{1,2}$ is the third term, $x_{3,1}$ is the forth term, $x_{2,2}$ is the fifth term, and so on.
A Bijection between this sequence and $\mathbb{N}$ can be seen because:

*

*For each of the elements of the union of the all the $E_n,\ $ for example $\ x_{3,1},\ $ the arrows do not go over this element twice and the arrows do not miss out any of these elements.

*In other words, for each element $x_{i,j}\ $ in Rudin's "arrow sequence": $x_{1,1},\ x_{2,1}\ x_{1,2}\ x_{3,1}\ x_{2,2}\ \ldots,\ $ there exists a unique positive integer $n$ such that $x_{i,j}\ $ is the $n$-th element of this sequence. Furthermore, for every integer $m,$ there exists a unique pair $(i,j)$ such that $x_{i,j}\ $ is the $m$-th element of the sequence. Thus the sequence is a bijection of $\mathbb{N}$.

A: The proof shows that there is a surjection from $\mathbb{N}\times\mathbb{N}$ to $S$(map each pair $(n,k)$ to $x_{nk}$), this means that there is an injection from $S\to \mathbb{N}\times\mathbb{N}$, since $\mathbb{N}\times\mathbb{N}$ is countable there is an injection $S\to T$ where $T$ is a subset of the naturals.
