How to prove that the limsup of a sequence is equal to its greatest subsequential limit? I have a very tricky problem that I'm having a hard time figuring out how to start. 
Basically, I want to prove that the supremum of the set of subsequential limits of a sequence is equal to the lim sup of the sequence.
So I have a sequence $S_n$. I want to show that its greatest subsequential limit (which could either be a real number, infinity, or negative infinity) is equal to the limit (as N goes to infinity) of the supremum of the set $X=\{S_n:n>N\}$, 
$$\lim_{N\to\infty} \sup \{S_n:n>N\}$$
which is the definition of limit superior. 
I'm having a hard time coming up with a way to go about this.

Use the definition: $\lim \sup S_n=\lim_{n\to \infty}T_n$ where $T_n=\sup \{S_m:m\geq n\}.$ Assume $-\infty< L=\lim \sup S_n<+\infty.$
(I). Take any $M>L.$ There are only finitely many $n$ such that $S_n\geq (M+L)/2.$ Because  otherwise $T_n\geq (M+L)/2$ for every $n,$ implying $\lim_{n\to \infty}T_n\geq (M+L)/2>L.$  
So $(S_n)_n$ cannot have a subsequence converging to $M$ for any $M>L.$ Because otherwise there are infinitely many $n$ with $|S_n-M|<(M-L)/2,$  implying there are infinitely many $n$ such that $S_n\geq (M+L)/2.$
(II). Take any $N<L.$  There do exist infinitely many $n$ such that $S_n>N.$ Otherwise $T_n\leq N$ except for finitely many $n,$ implying $\lim_{n\to \infty}T_n\leq N<L.$ 
(III). By (II) with $N =L-2^{-1},$ take $n(1)$ such that $S_{n(1)}> L-2^{-1}.$ Recursively, by (II) with $N=L-2^{-j-1}$  take $n(j+1)>n(j)$ such that $S_{n(j+1)}> L-2^{-j-1}.$
The sequence $(S_{n(j)})_j$ converges to $L.$
Because if $\epsilon >0,$ then by (I) with $M=L+\epsilon$, take $K\in \mathbb N$ large enough that $n\geq K\implies S_n<L+\epsilon.$ And  take $j_K\in \mathbb N$ large enough that $n(j_K)\geq K$ and $2^{-j_K}<\epsilon.$  Then $$j\geq j_K\implies |S_{n(j)}-L|<\epsilon.$$
Remark: $\lim \sup S_n$ is the the largest number $x$ such that, for every $r>0,$ the set $\{n: S_n\in [-r+x,r+x]\;\}$ is infinite.
