This is the explanation of the book, but for me I did not catch the idea of the set $v$, and I need a numerical example for it and the set $u_{k}$, could anyone help me please?
Is there a systematic way by which I can write $u_{k}$ so that $u_{k}$ give me all subsets of $\mathbb{N}$?
The idea is to construct a set which is none of $u_k$. This is done by assuring that $k \in v \leftrightarrow k\notin u_k$, this means that for any $k$ that $u_k\ne v$.
Note that the proof assumes there is an enumeration $u_k$ to start with and it shows that there is a $v\in 2^{\mathbb N}$ that is not part of the enumeration.
In fact the same argument is used to show that $2^X$ is larger than $X$. Which can be used to create an concrete example. For example assume that $X=\{0,1,2\}$ and suppose that we have an surjective mapping from $X$ to $2^X$ which would mean that we have three sets $u_0$, $u_1$ and $u_2$ which would make up $2^X$. Now take for example:
$$u_0 = \{1,2\}$$ $$u_1 = \{0,1\}$$ $$u_2 = \{0\}$$
Now we construct $v$ from this and we have $0\in v\leftrightarrow 0\notin u_0$, but $0\notin u_0$ so $0\in v$. In the same way we find that $1\notin v$ (because $1\in u_1$) and $2\in v$. So we have $v=\{0,2\}$ which is none of $u_0, u_1, u_2$.
What confuses you is possibly the appeal to contradiction, which is by no means necessary.
The book actually proves the following statement:
Suppose $u_1,u_2,\dots,u_n,\dotsc$ is a sequence of (pairwise distinct) subsets of $N$. Then there exists a subset $v$ of $N$ such that $v\ne u_n$, for every $n$.
The subset is built from the given sequence by $$ v=\{k: k\notin u_k\} $$ Indeed, let $n$ be a natural number: either $n\in v$ or $n\notin v$.
In the first case, $n\notin u_n$ by the very definition of $v$, so $v\ne u_n$; in the second case, $n\in u_n$, again by the very definition of $v$, so $v\ne u_n$. In both cases $v\ne u_n$.
In particular, no countable subset of $2^N$ can equal $2^N$, because from a countable subset we can produce an element of $2^N$ that is not in the given subset.
The more general statement that $|2^X|>|X|$, for every set $X$, can be proved similarly. Since, clearly, $|X|\le|2^X|$, the assert follows from
There is no surjective map $f\colon X\to 2^X$.
Indeed, suppose $f\colon X\to 2^X$ is any map. Define $$ V=\{x\in X:x\notin f(x)\} $$ Let us show that $V\ne f(x)$, for every $x$. Indeed, for $x\in X$ there are two cases
In the first case $x\notin f(x)$, so $V\ne f(x)$. In the second case, $x\in f(x)$, so $V\ne f(x)$.
Therefore $f$ is not surjective.
The idea of the proof, as you see, is the same (and goes back to Cantor himself).
