I am confused about the first sentence of this proof:
Theorem: Let $k$ be a positive integer. If $\{I_n\}$ is a sequence of $k$-cells such that $I_{n+1} \subset I_n$ for $n=1, 2, 3, \dots$, then $\cap_{1}^{\infty} I_n$ is not empty.
Proof: Let $I_n$ consist of all points $x=(x_1, \dots, x_k)$ such that $a_{n,j}\leq x_j \leq b_{n, j}$ for $1\leq j\leq k; n=1, 2, 3, \dots$ and put $I_{n,j}=[a_{n,j},b_{n,j}]$.
In the theorem, $k$-cell is defined as:
Given $a_i < b_i$, where $i=1, \dots, k$, $\{x=(x_1, \dots, x_k)\in R^k | a_i<x_i <b_i\}$ is called a $k$-cell.
My question is: What does it mean to "let $I_n$ consist of all points $x=(x_1, \dots, x_k)$ such that $a_{n,j}\leq x_j \leq b_{n, j}$ for $1\leq j\leq k; n=1, 2, 3, \dots$ and put $I_{n,j}=[a_{n,j},b_{n,j}]$?" I mean, doesn't $I_n$ contains $x$ naturally following from $a_{n,j}\leq x_j \leq b_{n, j}$ and the definition of "$k$-cells?" The order of how Rudin puts it just does not make sense to me. Any hint will be appreciated.
It seems to me that the given definition of a $k$-cell in your post is wrong. Consider $I_n$ to be the $k$-dimensional cube $$I_n =(0,\frac{1}{n})\times\cdots\times (0,\frac{1}{n})$$
Then $I_n$ is a $k$-cell according to your definition and $I_{n+1} \subset I_n$, but $\bigcap_{n=1}^{\infty}I_n = \emptyset$
Therefore, a $k$-cell in the theorem Rudin is trying to prove must be of this form:
$$A = \{(x_1,\cdots,x_k) \in \mathbb{R}^k: a_i \leq x_i \leq b_i\}$$
In other words, the intervals in a $k$-cell must be closed for the theorem to work.
I think Rudin's first paragraph is only telling you what a typical $k$-cell in his proof is assumed to look like. So, he's just fixing notations for the rest of the proof and you shouldn't spend too much time on it or over-complicate it.
In fact, what Rudin is trying to prove relies very much on the fact that $k$-cell, the way I defined it, is compact. If you don't know what compactness is, you should continue reading Rudin's Chapter 2.