Theorem 3.4.11 in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Limit superior and limit inferior of a bounded sequence of real numbers Here is Theorem 3.4.11 in the book Introduction to Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition: 

If $\left( x_n \right)$ is a bounded sequence of real numbers, then the following statements for a real number $x^*$ are equivalent. 
(a) $x^* = \limsup \left( x_n \right)$. 
(b) If $\varepsilon > 0$, there are at most a finite number of $n \in \mathbb{N}$ such that $x^* + \varepsilon < x_n$, but an infinite number of $n \in \mathbb{N}$ such that $x^* - \varepsilon < x_n$. 
(c) If $u_m = \sup \left\{ \ x_n \ \colon \ n \geq m \ \right\}$, then $x^* = \inf \left\{ \ u_m \ \colon \ m \in \mathbb{N} \ \right\} = \lim \left( u_m \right)$. 
(d) If $S$ is the set of subsequential limits of $\left( x_n \right)$, then $x^* = \sup S$. 

And, here is Definition 3.4.10: 

Let $X = \left( x_n \right)$ be a bounded sequence of real numbers. 
(a) The limit superior of $\left( x_n \right)$ is the infimum of the set $V$ of $v \in \mathbb{R}$ such that $v < x_n$ for at most a finite number of $n \in \mathbb{N}$. It is denoted by 
  $$ \limsup \left( x_n \right) \qquad \mbox{ or } \qquad \limsup X \qquad \mbox{ or } \qquad \overline{\lim} \left( x_n \right). $$
(b) The limit inferior of $\left( x_n \right)$ is the supremum of the set of $w \in \mathbb{R}$ such that $x_n < w$ for at most a finite number of $n \in \mathbb{N}$. It is denoted by 
  $$ \liminf  \left( x_n \right) \qquad \mbox{ or } \qquad \liminf X \qquad \mbox{ or } \qquad \underline{\lim} \left( x_n \right). $$

Finally, here is the proof of Theorem 3.4.11: 
[In what follows, I have included my questions and explanations within square brackets.] 

(a) implies (b). If $\varepsilon > 0$, then the fact that $x^*$ is an infimum implies that there exists a $v$ in $V$ such that $x^* \leq v < x^* + \varepsilon$. Therefore $x^*$ also belongs to $V$, [I think here it should be $x^*+\varepsilon$ instead of $x^*$. Am I right?] so there can be at most a finite number of $n \in \mathbb{N}$ such that $x^*+\varepsilon < x_n$. On the other hand, $x^*-\varepsilon$ is not in $V$ so there are an infinite number $n \in \mathbb{N}$ such that $x^* - \varepsilon < x_n$. 
(b) implies (c). If (b) holds, given $\varepsilon > 0$, then for all sufficiently large $m$ we have $u_m < x+\varepsilon$. [I think this should be $u_m < x^* + \varepsilon$. Am I right?] Therefore, $\inf \left\{ \ u_m \ \colon \ m \in \mathbb{N} \ \right\} \leq x^* + \varepsilon$. [I think we can even write this as $\inf \left\{ \ u_m \ \colon \ m \in \mathbb{N} \ \right\} < x^* + \varepsilon $. Am I right?] Also, since there are an infinite number of $n \in \mathbb{N}$ such that $x^* - \varepsilon < x_n$, then $x^* - \varepsilon < u_m$ for all $m \in \mathbb{N}$ and hence $x^* - \varepsilon \leq \inf \left\{ \ u_m \ \colon \ m \in \mathbb{N} \ \right\}$. Since $\varepsilon > 0$ is arbitrary, we conclude that $x^* = \inf \left\{ \ u_m \ \colon \ m \in \mathbb{N} \ \right\}$. Moreover, since the sequence $\left( u_m \right)$ is monotone decreasing, we have $\inf \left( u_m \right) = \lim \left( u_m \right)$. 
(c) implies (d). Suppose that $X^\prime = \left( x_{n_k} \right)$ is a convergent subsequence of $X = \left( x_n \right)$. Since $n_k \geq k$, we have $x_{n_k} \leq u_k$ and hence $\lim X^\prime \leq \lim \left( u_k \right) = x^*$. Conversely, there exists $n_1$ such that $u_1 - 1 \leq x_{n_1} \leq u_1$. [I think here we can even write $u_1 - 1 < x_{n_1} \leq u_1$. Am I right?] [Suppose that $n_k$ has been chosen.] Inductively choose $n_{k+1} > n_k$ such that 
  $$ u_k - \frac{1}{k+1} < x_{n_{k+1}} \leq u_k. $$
  [But I think we should write this as $u_{k+1} - \frac{1}{k+1} < x_{n_{k+1}} \leq u_{k+1}$ or as $u_{k+1} - \frac{1}{k+1} \leq x_{n_{k+1}} \leq u_{k+1}$. Am I right? What is an explicit procedure and justification of how $x_{n_2}$ can be chosen such that $n_2 > n_1$ and such that $u_2 - \frac{1}{2} \leq x_{n_2} \leq u_2$? I would appreciate a detailed account of this. ] Since $\lim \left( u_k \right) = x^*$, it follows that $x^* = \lim \left( x_{n_k} \right)$, and hence $x^* \in S$. [In fact, here we have even shown that $x^*$ is the largest element of $S$. Am I right?] 
(d) implies (a). Let $w = \sup S$. If $\varepsilon > 0$ is given, then [because there can be no subsequential limit of $\left( x_n \right)$ exceeding $w$ and, as there is always a convergent subsequence of every bounded sequence of real numbers, so there can be no subsequence of $\left( x_n \right)$ having infinitely many terms in the interval $(w, +\infty)$] there are at most finitely many $n$ with $w + \varepsilon < x_n$. Therefore $w+\varepsilon$ belongs to $V$ and $\limsup \left( x_n \right) \leq w+\varepsilon$. On the other hand, there exists a subsequence of $\left( x_n \right)$ converging to some number larger than $w - \varepsilon$, [We call that number $x$.] so that $w - \varepsilon$ is not in $V$,  [Is this because there are infinitely many terms of that convergent subsequence that are in any neighborhood of $x$ and hence are greater than $w + \varepsilon$?] and hence $w - \varepsilon \leq \limsup \left( x_n \right)$. [How is this true? Is it because of the fact that if $\limsup \left(x_n \right)$ were less than $w - \varepsilon$, then there would be some element $v \in V$ such that  $\limsup \left(x_n \right) \leq v < w - \varepsilon < x$ and by definition of set $V$ in that case there would be only finitely many terms of the sequence exceeding $v$ and hence only finitely many terms of the sequence in the neighborhood $( w - \varepsilon , 2x + w-\varepsilon )$, for example, of $x$?] Since $\varepsilon > 0$ is arbitrary, we conclude that $w = \limsup \left( x_n \right)$. 

In the above proof, I have included my questions and points of confusion within square brackets. Is there any occasion where (in any of my explanations within square brackets) I've gone wrong with my reasoning? 
In (c) implies (d), once we have shown that $x^*$ is an upper bound for the set $S$ of subsequential limits of $\left( x_n \right)$, is there an alternative approach to showing that $x^*$ is in fact the supremum of $S$ (or even an element of $S$)? 
 A: I will try to answer (c) implies (d). [All your queries in square brackets seem to echo mine. Since I have just started teaching myself limsup I take what in that section as gospel. I dare not ask...]
There exists $ n_1 $ such that
\begin{equation*}
 u_1 - 1 \leq x_{n_1} < u_1.
\end{equation*}
Choose $ n_2 > n_1 $, then
\begin{equation*}
 u_{n_1} - \frac{1}{2} \leq x_{n_2} < u_{n_1}.
\end{equation*}
Choose $ n_3 > n_2 $, then
\begin{equation*}
 u_{n_2} - \frac{1}{3} \leq x_{n_3} < u_{n_2}.
\end{equation*}
Inductively choose $ n_{k+1} > n_k $ such that
\begin{equation*}
 u_{n_k} - \frac{1}{k+1} \leq x_{n_{k+1}} < u_{n_k}.
\end{equation*}
Since $ u_1, u_{n_1}, u_{n_2}, \dotsc, $ is a subsequence of a convergent sequence $ (u_k) $, 
$ \lim_{k\to\infty} u_{n_k} = x^* $, then by Squeeze Theorem
$ \lim_{k\to\infty} x_{n_k} = x^* $.
Here we have shown there is a subsequence $ (x_{n_k}) $ with limit $ x^* $, and by the first part of the proof, all subsequences have limits less than $ x^* $. Therefore $ x^* $ is the supremum of all the subsequential limits.
It seems that I can get the desired result if I use the subsequence of $ (u_k) $, not the sequence itself, to construct the subsequence of $ (x_n) $ with the largest limit.
