Proof of Diagonal lemma 
Lemma :(Diagonal lemma)
Let $\{x_n\}_{n\in I_m}$ be a sequence $\forall m = 1, 2, ...$. Such that $\{x_n\}_{n\in I_{m+1}}$ is a subsequence of $\{x_n\}_{n\in I_m}$.
Then there exists a subsequence  $\{x_n\}_{n\in J}$ of $\{x_n\}_{n\in I_m}$ $\forall  m = 1, 2, ...$.

Proof:

Let  $J = \{n_1, n_2, ...\}$ such that : $n_m$ is the smallest member of
$$
I_1 \cap I_2 \cap.... \cap I_m \cap \{m,m + 1, ...\}
$$
Since $n_k \in I_m$ for all $k\ge m$, it follows that $\{x_n\}_{n\in J}$  is a sub-sequence of $\{x_n\}_{n\in I_m}$  for all $m = 1, 2, ...$

I read this proof several times, but I did not understand why :  $\{x_n\}_{n\in J}$  is a sub-sequence of $\{x_n\}_{n\in I_m}$.
Is the definition of a sub-sequence: the existence of an infinity of sub-sequence indices which are in the global sequence or else the sequence contains the entire sub-sequence?
An idea please
 A: Maybe I am missing the gist of this, but all this seems to be is an (unnecessarily?) abstruse formulation of the diagonalization procedure. Since
$I_{m+1}\subseteq I_m,$ we have  $I_1 \cap I_2 \cap.... \cap I_m \cap \{m,m + 1, ...\}=I_m\cap \{m,m + 1, ...\}$. 
So, the lemma provides the usual  recipe for choosing a subsequence which is $eventually$ a subsequence of each $\{x_n\}_{n\in I_m}:$
From the sequence $\{x_n\}_{n\in I_1}$, choose the least integer $n_1\in I_1$ and then the element of the sequence that corresponds to it, $x_{n_1}$. Now, from the sequence $\{x_n\}_{n\in I_2}$, choose, if this is possible, the least integer $n_2$ in $I_2$ that is greater than $n_1$ and then select the element $x_{n_2}\in \{x_n\}_{n\in I_2}$ that corresponds to it. If this is not possible, then the subsequence is the singleton $x_{n_1}$.
At the $k^{th}$ step, we have a sequence $\{n_j\}^k_{j=1}$ of integers, such that $n_{j+1}>n_j$ and such that the sequence $\{x_{n_1},\cdots, x_{n_k}\}$ satisfies: $\{x_{n_j}, x_{n_{j+1}},\cdots x_{n_k}\}$ is a subsequence of $\{x_n\}_{n\in I_j}$.
Now, from the sequence $\{x_n\}_{n\in I_{k+1}}$, if it is possible to choose the least integer $n_{k+1}\in I_{k+1}$, then do so and then select the element of the sequence that corresponds to it, $x_{n_{k+1}}$. If it is not possible to make this choice, then stop. This completes the induction step.
We now have either a finite sequence $\{x_{n_1},\cdots, x_{n_k}\}$ that satisfies: $\{x_{n_j}, x_{n_{j+1}},\cdots x_{n_k}\}$ is a subsequence of $\{x_n\}_{n\in I_j}$ or an infinite sequence with the same property. 
