Here is Prob. 10, Sec. 3.4, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Let $\left( x_n \right)$ be a bounded sequence and for each $n \in \mathbb{N}$ let $s_n \colon= \sup \left\{ \ x_k \ \colon \ k \geq n \ \right\}$ and $S \colon= \inf \left\{ s_n \right\}$. Show that there exists a subsequence of $\left( x_n \right)$ that converges to $S$.
My Attempt:
As $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S+1 > S$, so $S+1$ is not a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$, and thus there is a natural number $n_1$ such that $$ S \leq s_{n_1} < S+1. \tag{1} $$
Now as $ s_{n_1} = \sup \left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_1 \ \right\}$ and as $s_{n_1} - 1 < s_{n_1}$, so $s_{n_1} - 1$ is not an upper bound for the set $\left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_1 \ \right\}$, and thus we can find a natural number $k_1 \geq n_1$ such that $$ s_{n_1} - 1 < x_{k_1} \leq s_{n_1}. \tag{2} $$
From (1) and (2) we obtain $$ S - 1 \leq s_{n_1} - 1 < x_{k_1} \leq s_{n_1} < S+1, $$ and hence $$ S-1 < x_{k_1} < S + 1. \tag{3} $$
Again as $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S+1/2 > S$, so $S+1/2$ is not a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$, and thus there is a natural number $n_2$ such that $$ S \leq s_{n_2} < S+\frac{1}{2}. \tag{4} $$
Can we show whether $n_2 > n_1$? Or, is this condition really needed for our proof to go through?
Now as $s_{n_2} = \sup \left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_2 \ \right\}$ and as $s_{n_2} - 1/2 < s_{n_2}$, so $s_{n_2} - 1/2$ is not an upper bound for the set $\left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_2 \ \right\}$, and thus we can find a natural number $k_2 \geq n_2$ such that $$ s_{n_2} - \frac{1}{2} < x_{k_2} \leq s_{n_2}. \tag{5} $$
Can we show if $k_2 > k_1$? This condition is essential, isn't it?
From (4) and (5) we obtain $$ S - \frac{1}{2} \leq s_{n_2} - 1 < x_{k_2} \leq s_{n_2} < S+ \frac{1}{2}, $$ and hence $$ S - \frac{1}{2} < x_{k_2} < S + \frac{1}{2}. \tag{6} $$
Once we have shown that our $k_2$ is indeed greater than our $k_1$, we can repeat the above process and thus obtain a subsequence $\left( x_{k_m} \right)_{m \in \mathbb{N}}$ of $\left( x_n \right)$ such that $$ S - \frac{1}{m} < x_{k_m} < S + \frac{1}{m} \ \mbox{ for all } m \in \mathbb{N}. $$ And then our desired conclusion will follow. But how to accomplish this?
P.S.:
After reading through the answer below, I think I have the idea of how to proceed.
Here is my amended proof.
As $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S+1 > S$, so $S+1$ is not a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$, and thus there is a natural number $n_1$ such that $$ S \leq s_{n_1} < S+1. \tag{1} $$
Again as $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S-1 < S$, so $S - 1$ is also a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$, and thus for every natural number $n$ we have $$ S-1 < S \leq s_{n}, $$ and hence $$ S-1 < s_{n}. $$ In particular, for $n = n_1$, we can also write $$ S-1 < s_{n_1}. $$ And, as $s_{n_1} = \sup \left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_1 \ \right\}$ and as $ S-1 < s_{n_1}$, so $S-1$ is not an upper bound for the set $\left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_1 \ \right\}$ and thus there exists a natural number $k_1 \geq n_1$ such that $$ S-1 < x_{k_1} \leq s_{n_1}. \tag{2} $$
From (1) and (2) we obtain $$ S-1 < x_{k_1} \leq s_{n_1} < S+1, $$ and hence $$ S-1 < x_{k_1} < S+1. \tag{3} $$
Again as $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S+1/2 > S$, so $S+1/2$ is not a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and thus we can find a natural number $n_2$ such that $$ S \leq s_{n_2} < S+ \frac{1}{2}. \tag{4} $$
Again as $S = \inf \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and as $S-1/2 < S$, so $S-1/2$ is also a lower bound for the set $\left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\}$ and thus for every natural number $n$ we have $$ S - \frac{1}{2} < S \leq s_n, $$ which gives $$ S - \frac{1}{2} < s_n. $$ In particular, for $$r_2 \colon= \max \left\{ \ n_2, k_1 + 1 \ \right\}, \tag{A} $$ we obtain $$ S- \frac{1}{2} < s_{r_2}. $$
Now as $s_{r_2} = \sup \left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq r_2 \ \right\}$ and as $S-1/2 < s_{r_2}$, so $S - 1/2$ is not an upper bound for the set $\left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq r_2 \ \right\}$ and thus there exists a natural number $k_2 \geq r_2$ such that $$ S - \frac{1}{2} < x_{k_2} \leq s_{r_2}. \tag{5'} $$
As $r_2 \geq n_2$ [Note (A) above.], so the set $\left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq r_2 \ \right\} \subseteq \left\{ \ x_k \ \colon \ k \in \mathbb{N}, k \geq n_2 \ \right\}$ and so $$ s_{r_2} \leq s_{n_2}. \tag{5''} $$ From (5') and (5'') we obtain $$ S - \frac{1}{2} < x_{k_2} \leq s_{r_2} \leq s_{n_2}, $$ and hence $$ S-\frac{1}{2} < x_{k_2} \leq s_{n_2}. \tag{5} $$
From (4) and (5) we obtain $$ S - \frac{1}{2} < x_{k_2} \leq s_{n_2} < S+ \frac{1}{2}, $$ and hence $$ S - \frac{1}{2} < x_{k_2} < S+ \frac{1}{2}. \tag{6} $$
Finally as $k_2 \geq r_2$ and from (A) as $r_2 \geq k_1 + 1$, so $k_2 > k_1$ also. Thus our desired process can continue.
Denote $\limsup x_{n}=S$. There is an $n_{1}$ such that $$ S+1 > \sup_{k\geq n_{1}} x_{k} = s_{n_1}. $$ Since $$ S-1 < \inf_{n \in \mathbb{N} } \sup_{k \geq n} x_{k} \leq \sup_{k\geq n}x_{k}, $$ for each $n=1,2,...$, then $$ S-1 < \sup_{k\geq n_{1} } x_{k},$$ which implies then that there is an $k_{1} \geq n_{1}$ such that $$ S-1 < x_{k_{1}}.$$ We also have $$ x_{k_{1}} \leq \sup_{k\geq n_{1}}x_{k}<S+1. $$
There is an $n_{2}$ such that $$ S+1/2> \sup_{k\geq n_{2}} x_{k} .$$ Since $$ S-1/2 < \inf_{n \in \mathbb{N} } \sup_{k\geq n}x_{k} \leq \sup_{k\geq n}x_{k} $$ for each $n=1,2,...$, which implies then that $$ S-1/2 < \sup_{k\geq\max\{ n_{2}, k_{1}+1\}} x_{k}.$$ Choose a $k_2$ such that $$ k_{2} \geq \max\{n_{2}, k_{1}+1\} $$ and such that $$ S-1/2 < x_{k_{2}}. $$ We also have $$ x_{k_{2}} \leq \sup_{k\geq n_{2}} x_{k} < S+1/2. $$
Proceed in this way.