A subsequence of a convergent sequence converges to the same limit. Questions on proof. (Abbott p 57 2.5.1) Solutions to Homework 3 doesn`t duplicate. We have to prove that if $(a_{n})$ is a sequence in $\mathbb{R}$ with $\displaystyle \lim_{n\rightarrow\infty} a_n =a$, and if $(a_{n_{k}})_{k\in \mathbb{N}+}$ is a subsequence of $(a_{n})$ , then $\displaystyle \lim_{k\rightarrow\infty}a_{n_{k}}=a$.
We first need to know that $n_{k}\geq k$ for all $k\in Z_{>0}$. This is proved by induction on $k$. I omit this.
Let $e >0$. $\displaystyle \lim_{n\rightarrow\infty}a_{n}=a$ is posited, so there's $\color{violet}{N}\in \mathbb{N}$ such that for all $n\in\mathbb{N}$, $n \ge N \implies |a_n-a| < e$.
Let $k\in \mathbb{N}$ with $k\geq N.$ Then ${n_k}\geq k\geq N$. Therefore $|a_{n_{k}}-a|<e$.

I know we must find $ N\in\mathbb{N}$ such that  $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.
  As the proof overhead shows, this $N$ is the same as the posited $\color{violet}{N}$. But what engenders $(♫)$ ?
Is proof saying $n \ge N \implies |a_n-a| < e$
   implies $\color{red}{k} \ge N \implies |a_\color{red}{k}-a| < e$
   implies $(♫)$, because $n_k \ge k \ge N$?
If yes, then I don't understand how $k$ can be replaced by $n_k$? 


Right this is what is going on. 
We first assume $ e\gt 0 $ is an arbitrary positive quantity. You've clearly understood that we need to produce a natural number $N$ such that for every element $a_{n_k}$  after the $N$th term of the sequence $(a_{n_k})$ we have $|a_{n_k} - a| \lt e$. 
What we know is that (since $a_n \to a$) there is a a natural number $N$ such that every term in the sequence $(a_n)$ past the $N$th term is such that $|a_n - a| \lt e$. Now what is being argued is that if $a_{n_j}$ is a term in the subsequence $(a_{n_k})$ beyond "its" $N$th term then since $a_{n_j}$ is also a part of the original sequence and since $\{n_k\}$ is a strictly increasing sequence of natural numbers, $a_{n_j}$ is surely beyond the $N$th term in the original sequence and hence the inequality holds. This is what is projected by $ k \le n_k $. 
