I'm having trouble with a definition of the upper and lower limits, and a theorem that follows it. The following is the definition.

Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ such that $s_{n_{k}}\rightarrow{x}$. This set $E$ contains all subsequential limits, plus possibly the numbers $+\infty$, $-\infty$. Now, putting $s^* = \sup (E)$ and $s_{*} = \inf (E)$. The numbers $s^{*}$ and $s_{*}$ are called upper and lower limits of ${s_n}$. $${\lim_{n \to \infty}} \sup s_n = s^{*}$$ $${\lim_{n \to \infty}}\inf s_n = s_{*}$$

I have no idea what this means.
From my understanding, if a sequence has a limit the limit is unique. Why is this definition implying that a sequence has multiple limits? 
Or is it implying that a subsequence of a sequence can have a different limit ?
My book lacks in examples and I cannot figure out what's going on at all ...
Also, I learned that $\infty$ is not a number. Why is this definition treating it as if it is one ?

Theorem. Let ${s_n}$ be a sequence of real numbers. Then $s^*$ has the following properties.
  a) $s^* \in E$
  b) If $x>s^*$, $\exists N \in \Bbb Z$ such that $\forall n \ge N$, $s_n < x$.
Moreover, $s^*$ is unique. 

I was able to understand the proof of a), and b), but I couldn't really understand the proof of the uniqueness. The book says,

Suppose $p<q$ where both $p$ and $q$ are upper limits. Choose $x$ such that $p < x < q$. Since $p$ satisfies b), we have $s_n<x$ for $n \ge N$. But then $q$ cannot satisfy a).

1) Why is it guaranteed that such $x$ between $p$ and $q$ exist?
2) Why will $q$ not satisfy a) ?


Why is this definition implying that a sequence has multiple limits?

Andres has given a good answer to this in the comments. In short, even if a sequence has no unique limit, subsequences of the sequence can have a unique limit. The quoted definition is talking about limits of subsequences of a sequence.

Why is it guaranteed that such x between p and q exist?

We assumed that $p<q$, this means that $q-p=\varepsilon$ for some $\varepsilon>0$. You can take $x$ to be $p+\frac{\varepsilon}{2}$.

Why will q not satisfy a) ?

Since $s_n<x<q$ for $n \ge N$,
$${\lim_{n \to \infty}} \sup s_n \le x<q.$$ 
So $q$ can't be the supremum of $s_n$ as $n\rightarrow \infty$ because $x$ will always be a tighter bound.
