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 $n_k$? 


I'm going to post a separate answer because I think I can be clearer than the accepted answer.

I know we must find $ N\in\mathbb{N}$ such that  $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.    

No, this is not what you need to show!  You need to show that there exists $N$, such that for all $\boldsymbol{k} \ge N, |a_{n_k} - a| < \epsilon$.
This is because $\boldsymbol{a_{n_k}}$ is a sequence in $\boldsymbol{k}$, not a sequence in $\boldsymbol{n}$.  It takes in an integer $\boldsymbol{k}$ and gives you an integer in return.

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 $n_k \ge k \ge N$?

Here is what the proof is saying.  Fix any $k \ge N$.
Then $n_k \ge k \ge N$.
Thus $|a_{n_k} - a| < \epsilon$, Q.E.D.
If you have had trouble understanding any of the English in this answer, please let me know and I will try to clarify.
