Here is Prob. 14, 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 let $s \colon= \sup \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$. Show that if $s \not\in \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$, then there is a subsequence of $\left( x_n \right)$ that converges to $s$.
My Attempt:
As $s-1 < s$, so there exists a natural number $n_1$ such that $$ s-1 < x_{n_1} \leq s. $$ But $s \not\in \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$ and $x_{n_1} \in \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$. So we must have $$ s-1 < x_{n_1} < s. \tag{1} $$
As $s-1/2 < s$, so there exists a natural number $n_2$ such that $$ s - \frac{1}{2} < x_{n_2} \leq s. $$ But again $s \not\in \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$ and $x_{n_2} \in \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\}$. So we must have $$ s- \frac{1}{2} < x_{n_2} < s. \tag{2} $$
If there were no $n_2 > n_1$ for which (2) can hold, then we would obtain $$ x_n \leq s- \frac{1}{2} \ \mbox{ for every natural number } n > n_1. $$ So, for every natural number $n$, we would obtain
$$ x_n \leq \max \left\{ \ s- \frac{1}{2}, x_1, \ldots, x_{n_1} \right\} < s. $$ And this would then imply that $$ \sup \left\{ \ x_n \ \colon \ n \in \mathbb{N} \ \right\} < s. $$So there does exist a natural number $n_2 > n_1$ for which (2) can hold.
Now suppose that $n_1, n_2, \ldots, n_k$ have been found such that $n_1 < n_2 < \cdots < n_k$ and such that $$ s - \frac{1}{j} < x_{n_j} < s \ \mbox{ for all } j = 1, \ldots, k. $$ Then just as we proceeded from $n_1$ to $n_2$, we can find a natural number $n_{k+1} > n_k$ such that $$ s - \frac{1}{k+1} < x_{n_{k+1} } < s. $$
In this way we obtain a subsequence $\left( x_{n_k} \right)$ of $\left( x_n \right)$ that satisfies $$ s - \frac{1}{k} < x_{n_k} < s $$ for all $k \in \mathbb{N}$ and hence converges to $s$.
Is this proof correct in each and every one of its steps? If not, then where is it incorrect? Is the presentation clear enough too?
Your proof is correct and clear! As you show in your proof, we want to define our subsequence inductively:
Calling $\sup\{x_n: n\in\mathbb{N}\}=s$, we know that there must exist a term of the sequence, $x^*_n$, satisfying $s-1<x^*_n<s$. Let this be the first term of our subsequence: $x_{n_1}=x^*_n$.
Now, given $x_{n_k}$ for some $k\geq 1$, choose $x_{n_{k+1}}$ so that (i) $n_{k+1}>n_k$ and (ii) $s-\frac{1}{k+1}<x_{n_{k+1}}<s$.
Condition (i) ensures that the new sequence we're constructing is indeed a subsequence of $(x_n)$. Condition (ii) ensures that we have $$|x_{n_k}-s|<\frac{1}{k}$$ for all $k\in\mathbb{N}$, so that the subsequence has the desired property, namely, that $x_{n_k}\longrightarrow s$.