I want to prove that a sequence of real numbers $\{s_n\}$ converges to $s$ if and only if $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$.
Here are my definitions:
For any sequence of real numbers $\{s_n\}$, let $E$ be the set of all subsequential limits of $\{s_n\}$, including possibly $+\infty$ and/or $-\infty$ if any subsequence of $\{s_n\}$ diverges to infinity. Then $\limsup_{n \to \infty} s_n = \sup E$, and $\liminf_{n \to \infty} s_n = \inf E$.
I know the theorem that a sequence converges to a point if and only if every one of its subsequences converge to that same point, so one direction of this proof is easy:
If $\{s_n\}$ converges to some point $s \in \mathbb{R}$, then every subsequence of $\{s_n\}$ converges to $s$. So the set $E$ of every subsequential limit of $\{s_n\}$ consists of the single point $s$, so
$$\limsup_{n \to \infty} s_n = \sup \{s\} = s = \inf \{s\} = \liminf_{n \to \infty} s_n$$
But the other direction seems more tricky…
If $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$, then every convergent subsequence converges to the same point $s$. Also, there can be no subsequences which diverge to infinity (otherwise $\limsup_{n \to \infty} s_n$ would be $+\infty$, or $\liminf_{n \to \infty} s_n$ would be $-\infty$).
But can't there be subsequences which diverge otherwise? And wouldn't that throw off the convergence of $\{s_n\}$?
EDIT:
I'd also be willing to accept a solution which makes use of the "Pinching Theorem" (if $a_n \leq s_n \leq b_n$ for every $n \in \mathbb{N}$, and if $a_n \to s$ and $b_n \to s$, then $s_n \to s$).
Best Answer
Useful facts that you should verify:
Any unbounded sequence has a subsequence diverging to $\infty$ or to $-\infty$.
Any bounded sequence has a convergent subsequence.
You correctly point out that the hypothesis that $\limsup_{n \to \infty} s_n$ and $\liminf_{n \to \infty} s_n$ are both finite implies that $(s_n)_{n=1}^{\infty}$ has no subsequences that diverge to infinity. But (1) implies that more is true: the sequence $(s_n)_{n=1}^{\infty}$ must be bounded.
Fix any $\epsilon > 0$.
There cannot be infinitely many $n$ for which $s_n \geq s + \epsilon$, because you could select out of them a subsequence $y_k = s_{n_k}$ satisfying $y_k \geq s + \epsilon$ for all $k$.
Similarly, there cannot be infinitely many $n$ for which $s_n \leq s - \epsilon$.
So there is a positive integer $N$ with the property that whenever $n \geq N$ one has $$ s - \epsilon < s_n < s + \epsilon, $$ and since $\epsilon > 0$ was arbitrary, $(s_n)_{n=1}^{\infty}$ converges to $s$.