Rudin's Theorem 3.17 Uniqueness of $\limsup s_n$ 
Let $\{s_n \}$ be a sequence of real numbers. Let $E$ be the set of all subsequential limits of $\{s_n \}$ and $s^* = \sup E = \limsup\limits_{n\to\infty}s_n$. Then $s^*$ has the following two properties:
(a) $s^* \in E$.
(b) If $x> s^*$, there is an integer $N$ such that $n \geq N$ implies $s_n < x$.
Moreover, $s^*$ is the only number with the properties (a) and (b).
Of course, an analogous result is true for $s_*$.

Here's Rudin's proof:

(a) If $s^* = +\infty$, then $E$ is not bounded above; hence $\{s_n\}$ is not bounded above, and there is a subsequence $\{s_{n_k}\}$ such that $s_{n_k} \rightarrow +\infty$.
If $s^*$ is real, then $E$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.
If $s^* = -\infty$, then $E$ contains only one element, namely $-\infty$, and there is no subsequential limit. Hence, for any real $M$, $s_n > M$ for at most a finite number of values of $n$, so that $s_n \rightarrow -\infty$.
This establishes (a) in all cases.
(b) Suppose there is a number $x > s^*$ such that $s_n \geq x$ for infinitely many values of $n$. In that case, there is a number $y \in E$ such that $y \geq x > s^*$, contradicting the definition of $s^*$.
Thus $s^*$ satisfies both (a) and (b).
To show the uniqueness, suppose there are two numbers, $p$ and $q$, which satisfy (a) and (b), and suppose $p < q$. Choose $x$ such that $p < x < q$. Since $p$ satisfies (b), we have $s_n < x$ for $n \geq N$. But then $q$ cannot satisfy (a).

In the uniqueness part, I can't see why the last statement implies that $q$ cannot satisfy (a). I tried proving by contradition but getting nowhere.
 A: Suppose there are two numbers, $p$ and $q$, which satisfy (a) and (b), and suppose $p<q$. Choose $x$ such that $p<x<q$. Since $p$ satisfies (b), we have $s_n<x$ for all $n≥N$. This means that for $n≥N$, $|q-s_n|\geq |q-x|$. But then $q$ cannot be a subsequential limit. 
Contradiction.
A: A more detailed answer:
Suppose that there are two numbers, $p$ and $q$, which satisfy $(a)$ and $(b)$. (Notice that by "numbers", we allow $p$, $q$ to be $+\infty$, $-\infty$, or real numbers.) Without loss of generality, let $p < q$. Then choose a real number $x$ such that $p <x< q$ (This is always possible). Since $p$ satisfies $(b)$, we know there exists an integer $N$ such that $n \geq N$ implies $s_{n} < x$. Then $q$ cannot be in $E$.
To see the last argument, suppose we have $q \in E$. Notice that $\{s_{n}\}$ is bounded above by $\max \{s_{1}, s_{2}, \dots, s_{N}, x\}$, then we cannot have a subsequence of $\{s_{n}\}$ that goes to $+\infty$. In other words, $q \ne +\infty$. It follows that $q$ can only possibly be a real number ($q \ne -\infty$ since $q > x$ by assumption). But we cannot have a subsequence converges to a real $q$ also. (Since there is a $N$ after which all the $s_{n}$'s are less than $x$. Suppose there is a subsequence converging to $q$, then let $\delta \in (0, q-x)$, then there is an $M$ such that for all $n_{k} \geq M$, $|s_{n_{k}}-q| < \delta$, by definition. But that implies $s_{n_{k}} > x$ if $n_{k} \geq M$, a contradiction.) Thus we cannot have $q \in E$, which finishes our proof.
