Can anyone please clarify for me certain points I have in the proof of Theorem 3.17 in Baby Rudin?

I'm currently studying, or revising, Walter Rudin's Principles of Mathematical Analysis, 3rd eition, and I'm stuck with the proof of Theorem 3.17.

Here's the statement of the Theorem:

Let $\{s_n\}$ be a sequence of real numbers, let $E$ be the set of all subsequential limits of $\{s_n\}$ in the extended real number system, and let $$s^* \colon= \sup E.$$ That is, let $s^*$ be the upper limit (or limit superior) of the sequence $\{s_n\}$. Then

(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).

I've figured out that if $s^* = +\infty$, then the set $E$ of subsequential limits is unbounded above in $\mathbb{R}$; so for every positive integer $N$ there is an element $x^{(N)}$ of $E$ such that $x^{(N)} > N$. Now since $x^{(N)}$ is the limit of a subsequence, say $\{s_{\phi_N(n)}\}$, of the sequence $\{s_n\}$, so there is a natural number $n_N$ such that $s_{n_N} > N$. So the subsequence $\{s_{n_N}\}$ diverges to $+\infty$. Am I right?

If $s^*$ is real, then the set $E$ is bounded above, but how to convince ourselves that in this case the set $E$ is NON-EMPTY in the first place?

And, if $s^* = -\infty$, then the set $E$ has no real number as an element; so that the only possible element of $E$ is $-\infty$. Now how to go about showing that this IS actually the case?

Finally, how to show that if $x > s^*$ and if $s_n \geq x$ for infinitely many $n$, then there exists an element $y$ in $E$ such that $y > s^*$?

1. Yes. You should find a divergent subsequence so that $+\infty \in E$.
2. We can show that $E$ cannot be empty. If $\{s_n\}$ is unbounded above or below, then $+\infty \in E$ or $-\infty \in E$ for reasons similar to 1. If $\{s_n\}$ is bounded, then by Bolzanno Weierstrass there must be a convergent subsequence, so $E$ again is nonempty.
3. That simply follows from the definition. If $s^\ast = -\infty$ is the supremum of $E$, then the existence of any real number in $E$ would make $s^\ast$ not an upper bound. Contradiction.
4. Take $\{s_{k_n}\}$ be the subsequence of all elements at least $x$. By 2., the set of subsequence limit points of $\{s_{k_n}\}$, call it $A \subset E$ is nonempty. Let $y \in A$. Since $s_{k_n} \geq x$ for all $n$, $y$ cannot be less than $x$. Thus $y \geq x > s^\ast$.