Theorem 11.2.i; Elementary Analysis, Ross 2nd Ed I am having trouble understanding what this theorem means:
11.2 Theorem
Let $(s_n)$ be a sequence. 
i. If t is in $\mathbb{R}$, then there is a subsequence of $(s_n)$ converging to t if and only if the set $\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$ is infinite for all $\epsilon > 0.$
Does this mean: 
If the $lims_n =t$ for $n\geq N$ and the set of all n's, for which this is true, is infinite,  then $\exists (s_{{n}_{k}})$ such that $(s_{{n}_{k}}) \rightarrow t$ (i.e. I can pick some n's from my infinite set to create indices of a subsequence that will converge to t)?
&
Suppose I have some $(s_{{n}_{k}}) \rightarrow t$, then there exists an infinite set of n's such that $\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$ 
(i.e. If I can find a subsequence, from the given parent sequence, that converges to t, then there must exist an infinite set of indices that create a sequence convergent to t)?
Particularly, I am confused about the usage of infinite and finite sets in both cases. I am interpreting that there are infinite subsequences that converge... 
 A: 
Suppose I have some $(s_{{n}_{k}}) \rightarrow t$, then there exists an infinite set of n's such that $\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$
(i.e. If I can find a subsequence, from the given parent sequence, that converges to t, then there must exist an infinite set of indices that create a sequence convergent to t)?

This is correct. Note how this is just the forward implication of the if-and-only-if statement.

Let $(s_n)$ be a sequence.
i. If t is in $\mathbb{R}$, then there is a subsequence of $(s_n)$ converging to t if and only if the set $\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$ is infinite for all $\epsilon > 0.$

You also say

If the $lim\ s_n =t$ for $n\geq N$ and the set of all n's, for which this is true, is infinite,  then $\exists (s_{{n}_{k}})$ such that $(s_{{n}_{k}}) \rightarrow t$ (i.e. I can pick some n's from my infinite set to create indices of a subsequence that will converge to t)?

This is where I think you're confusing things. You say that $\lim_{n \to \infty} s_n = t $, but the theorem is general, it does not require that $s_n$ should converge. It holds for any $s_n$.
I think where you are confused are is that you are thinking that the statement

$\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$ is infinite for

is saying that after a $N$, all elements of the sequence must follow this. But it is actually saying that $|s_{n}-t|<\epsilon$ must be true for infinite values of $n$, irrespective of where these elements are in the sequence.
For example, consider the sequence $<s_n> = <t_1, u_1, t_2, u_2, ....>$ where $t_n \to t$ and $u_n \to u, u \neq t$. Then there is no $n \geq N$ for which $\{|s_{n}-t|<\epsilon \}$ or $\{|s_{n}-u|<\epsilon \}$ is true, yet there exist two subsequences of $s_n$ which converge to limits $t,u$ respectively.
In the context of this example, what the backward implication of the theorem is saying is that it is possible that there is a subsequence of $s_n$ which converges to $t$ only because there are infinite terms of $s_n$ which satisfy $\{|s_{n}-t|<\epsilon \}$ and similarly for $u$.
That is, in general, if the set $\{n \in \mathbb{N}: |s_{n}-t|<\epsilon \}$ is infinite for all $\epsilon > 0$ then there must be a subsequence of $s_n$ which converges to $t$. If it is finite then there cannot exist a subsequence of $s_n$ which would converge to $t$.
