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.

You have to prove it in two steps.
First, prove that the greatest sequential limit cannot be greater than the limsup (that's easy, using reductio ab absurdum, suppose a sequential limit is greater than the limsup, and derive a contradiction).
Then, prove that the greatest sequential limit cannot be smaller than the limsup (explicitly build a sequence whose limit is greater than any real number strictly smaller than the limsup).
The only other option is that those two limits are equal.
