Proof of convergent Cauchy Sequence, if sime subsequence converges to p 
The idea is clear. But can you explain me please in more detail the procedure of chosing  $N$ for our subsequence. Why $N$ must be greater or equall $N_1$?
And why  $n_{N+1} >N+1 >N_1$?
 A: The $N_1$ is chosen to be greter than or equal to $N$ so that the condition for the sequence $p_n$ to be Cauchy is satisfied. Similarly $n_{N+1}\geq N+1>N_1$ because $n_k$ being a subsequence of $n_{n\in\mathbb{N}}$, the $kth$ term of the subsequence should necessarily be greater than or equal to the $kth$ term of the sequence itself, because the sequence $(n)_{n\in \mathbb{N}}$ is increasing
A: The rank for which your sequence has consecutive terms very close is $N_1$, that means after $N_1$ (ie: for $n\geq N_1$), all terms are close to each other.
Also you know that after $N$, all the terms of your subsequence and $p$ are close.
If you want both properties to be true, you must choose $n$ greater than $N_1$ and $N$.
For your last question, remember that an extraction $\phi$ is such that $\phi(n)>n$. In this redaction, $n_k > k$.
In this redaction, the author made the choice to directly take $N$ greater than $N_1$, since it's true for all rank greater than $N$, you must always consider it greater than $N$.
