I want to check my proof for this question:
Suppose a sequence {$a_n$} does not converge to 17. Prove that there exists some $\epsilon$ > 0 and a subsequence {$a_{nk}$} so that $|a_{nk} - 17|$ > $\epsilon$ for all $k$.
My solution: Let $\epsilon$ > 0. Since {$a_n$} does not converge to 17, there is some $N$ so that $n$ > $N$,$| a_{n} - 17| \geq \epsilon$.
We'll use the same $N$ to prove $a_{nk}$ does not converge to 17.
Suppose $k$ > $N$. But $n\geq k$. So $n_k\geq k > N$. So $n_k$ > $N$.
Hence $|a_{nk} - 17|$ > $\epsilon$ for all $k$. So {$a_{nk}$} does not also converge to 17.
The elements of a proof are there. We reorganize them so that they are clearer.
"The sequence $(a_n)$ converges to $17$" means that given any $\epsilon \gt 0$, there is an $N$ such that for all $n\ge N$, we have $|a_n -17| \lt \epsilon$.
So the fact that the sequence does not converge to $17$ means that there is an $\epsilon_0$ so that no matter how big $N$ is, there is an $n\ge N$ such that $|a_n-17\ge \epsilon_0$. The preceding sentence is the key to the whole thing.
Let $\epsilon_0$ be a number guaranteed to exist by the fact the sequence does not converge to $17$. Let $\epsilon=\epsilon_0/2$. This $\epsilon$ will be fixed for the remainder of the argument. So no matter how big $N$ is, there is an $n\ge N$ such that $|a_n-17|\gt \epsilon$.
Let $N=1$. Because of the failure of the sequence to converge to $17$, there is an $n_1\ge N$ such that $|a_{n_1}-17|\gt \epsilon$.
Recall that no matter how big $N$ is, there is an $n\ge N$ such that $|a_n-17|\gt \epsilon$. Let $N=n_1+1$. There is an $n_2\ge n_1+1$ such that $|a_{n_2}-17|\gt \epsilon$.
Let $N=n_2+1$. There is an $n_3\ge N$ such that $a_{n_3}-17|\gt \epsilon$.
And so on.
In this way, we can produce an infinite sequence $n_1\lt n_2\lt n_3\lt \cdots$ of indices such that $|a_{n_k}-17|\gt \epsilon$ for all $k$. That gives us the subsequence of $(a_n)$ asked for.
Remarks: $1.$ If we want to be quite formal about things, we could say suppose we have defined $n_1,n_2, n_k$ such that $n_1\lt n_2\lt \cdots\lt n_k$ such that $|a_{n_i}-17|\gt \epsilon$ for $1\le i\le k$. We define $n_{k+1}$ as the smallest index $j\gt n_k$ such that $|a_j-17|\gt \epsilon$.
$2.$ The $\epsilon_0$ to $\epsilon=\epsilon_0/2$ bit may have seemed mysterious. That was just to turn the $\ge \epsilon_0$ that we get by negating convergence to a $\gt \epsilon$.