Proving that a sequence converges This is what I've tried:
(a) Using mathematical induction, I can show that $b_{i+1}$ is smaller than $b_i$.
Therefore, $b_i$ is monotonically decreasing.
Then, I thought it's kind of obvious that $b_i$ is bounded because $\{a_i\}$ is a bounded sequence in $\mathbb{R}$.
Is this correct?
(b) Since $b_i$ converges, I can say $\lim _{i\to \infty} b_i  = p$ for some $p \in\mathbb{R}$. And this $p$ will equal to $\sup a_k$ when $k$ is large enough.
Hence, by definition, (b) is proven. Is this correct?
Thank you so much!

*I added the definition of lim sup below.

+Theorem 3.17

 A: (a) You are right, you can formalise your second sentence by applying the monotone convergence theorem.
(b) As you point out, this follows just from the definition of the limit superior and of $\{b_n\}$,
$$\limsup_{n\rightarrow\infty}a_n:=\lim_{n\rightarrow\infty}\left(\sup_{m\geq n}a_m\right)=\lim_{n\rightarrow\infty} b_n.$$
EDIT 2: In what follows I am assuming that $(*)$ is finite. However, if $(*)=+\infty$ then one can re-do the proof by following the same steps.
EDIT: We can show that both definitions are equivalent, that is
$$\lim_{n\rightarrow\infty}\left(\sup_{m\geq n}a_m\right)=\sup\{x:\exists\{a_{n_k}\}\subseteq\{a_n\},a_{n_k}\rightarrow x\text{ as }k\rightarrow\infty\},\quad\quad (*)$$
simply by using the triangle inequality and the definitions of the supremum of a set and the limit of a sequence. We do this in two steps:

By the definition of the supremum, we have that
$$\sup\{x:\exists\{a_{n_k}\}\subseteq\{a_n\},a_{n_k}\rightarrow x\text{ as }k\rightarrow\infty\}=p$$
implies that for every $\varepsilon>0$ there exists a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ such that 
$$\left|\lim_{k\rightarrow\infty}a_{n_k}-p\right|\leq \varepsilon.$$
By the definition of limit, we have that there exists a $K$ such that 
$$\left|a_{n_k}-\lim_{k\rightarrow\infty}a_{n_k}\right|\leq \varepsilon,$$
for all $k\geq K$. In particular, we have that for every $n\geq0$ we can find an $N$ (just pick any in $\{n_k\geq n:k\geq K\}$) such that 
$$\left|a_{N}-p\right|= \left|a_{N}-\lim_{k\rightarrow\infty}a_{n_k}+\lim_{k\rightarrow\infty}a_{n_k}-p\right| =\left|a_{N}-\lim_{k\rightarrow\infty}a_{n_k}\right|+\left|\lim_{k\rightarrow\infty}a_{n_k}-p\right|\leq2\varepsilon$$
So, for any $n$, pick the corresponding $N$ as above, and we have that
$$\sup_{m\geq n}a_m\geq a_N\geq p-2\varepsilon.$$
Since the $\varepsilon$ was arbitrary we have that for any $n$, $\sup_{m\geq n}a_m\geq p$. Hence,
$$\lim_{n\rightarrow\infty}\left(\sup_{m\geq n}a_m\right)\geq p=\sup\{x:\exists\{a_{n_k}\}\subseteq\{a_n\},a_{n_k}\rightarrow x\text{ as }k\rightarrow\infty\}.$$

We can establish the other inequality in a very similar fashion. Let
$$\lim_{n\rightarrow\infty}\left(\sup_{m\geq n}a_m\right)=q.$$
Then, for each $k>0$ we can find an $m_k$ such that 
$$\left|\sup_{m\geq m_k}a_m-q\right|\leq \frac{1}{2k}.$$
Similarly, we can find an $n_k\geq m_k$ such that 
$$\left|a_{n_k}-\sup_{m\geq m_k}a_m\right|\leq\frac{1}{2k}.$$
So, 
$$\left|a_{n_k}-q\right|\leq \left|a_{n_k}-\sup_{m\geq m_k}a_m+\sup_{m\geq m_k}a_m-q\right|\leq \left|a_{n_k}-\sup_{m\geq m_k}a_m\right|+\left|\sup_{m\geq m_k}a_m-q\right|\leq \frac{1}{k}. $$
Hence, we have constructed a subsequence $\{a_{n_k}\}$ such that $a_{n_k}\rightarrow q$ as $k\rightarrow\infty$. Thus,
$$\sup\{x:\exists\{a_{n_k}\}\subseteq\{a_n\},a_{n_k}\rightarrow x\text{ as }k\rightarrow\infty\}\geq q =\lim_{n\rightarrow\infty}\left(\sup_{m\geq n}a_m\right).$$
Putting together both inequalities we get $(*)$.
A: You're defining $$\ell^*=\limsup\limits_{n\to\infty}s_n$$ as the greatest upper bound of all subsequential limits of $\langle s_n\rangle$. We want to prove that, $\lim\limits_{n\to\infty}x_n=\ell^*$.
First, we show 

LEMMA1 Let $\langle s_n\rangle$ be a sequence. Then there exists a subsequence $n_k$ such that $$\lim\limits_{k\to\infty}s_{n_k}=\ell^*=\sup S$$ where $S=\{x:x\text{ is a subsequential limit of }s_n\}$.

P If $\ell^*=+\infty$ then the sequence is unbounded and we can obtain a sequence $n_k$ such that $s_{n_k}>k$ and the claim is true. Now suppose the sequence is bounded. Then $\ell^*$ exists finitely. By definition, given $N>0$ there exists $x\in S$ such that $$x \leq \ell^*<x+\frac 1{2N}$$
Thus $0 \leq x-\ell^*<\dfrac 1{2N}$ so we can say $|x-\ell^*|<\dfrac 1{2N}$. Let $j_k$ be such that $$s_{j_k}\to x$$
Then there exists $M=M_N$ such that implies $$|s_{j_M}-x|< \dfrac 1{2N}$$
But then $$\left| {{\ell ^*} - {s_{{j_M}}}} \right| \leqslant \left| {{\ell ^*} - x} \right| + \left| {x - {s_{{j_M}}}} \right| < \frac{1}{2N} + \frac{1}{2N} = \frac 1N$$
Now let $N$ vary throughout $1,2,\ldots$. We construct a sequence $M_1,M_2,\ldots$ as follows. Let $M_1$ be the number promised such that $$\left| {\ell ^*} - s_{m_{1,M_1}} \right| \leqslant 1$$ where $s_{m_{1,k}}\to x_1$ for some $x_1\in S$ as before. Now let $M_2$ be first number such that $m_{2,M_2}>m_{1,M_1}$ and $$\left| {{\ell ^*} - {s_{{m_{{2,M_2}}}}}} \right| \leqslant \frac{1}{2}$$ where $s_{m_{2,k}}\to x_2$ for some $x_2\in S$. Continue like this: we obtain a sequence $n_k=m_{k,M_k}$ such that $$\left| {{\ell ^*} - {s_{{n_k}}}} \right| \leqslant \frac{1}{k}$$ which means $$\mathop {\lim }\limits_{k \to \infty } {s_{{n_k}}} = {\ell ^*}$$ $\hspace{15.5 cm} \blacktriangle$
By the same token, we can show 

LEMMA2 There exists a subsequence such that $s_{m_k}\to \lim\limits_{n\to\infty}x_n$. I leave the proof to you, since it is very similar to the above.


Now all is clear to prove

PROP Let $\langle s_n\rangle $ be a sequence. Define $x_n=\sup\limits_{k\geq n}\{s_k\}$ and$\lim\limits_{n\to\infty}x_n=\ell$. Then $\ell=\ell^*$

P By the previous LEMMA there exist subsequences such that $$\lim\limits_{k\to\infty}s_{n_k}=\ell^*\\ \lim\limits_{j\to\infty}s_{m_j}=\ell$$
By definition of $\ell^*$, we must have $\ell\leq \ell^*$. It remains to show $\ell^*\leq \ell$.  Pick $n>0$. By the very definition of the $\langle x_n\rangle $ we must have $$x_n\geq s_k$$ for each $k\geq n$. In particular $$x_n\geq s_{n_j}$$ for each $n_j>n$. This means that $$x_n\geq \ell^*$$
Since $n$ was arbitrary, $x_n\geq \ell^*$ for each $n$, so $\ell\geq \ell^*$ $\hspace{15.5 cm} \blacktriangle$
