Source: Elementary Analysis By Kenneth A. Ross
I am having trouble understanding the last part of the proof of theorem 11.7.
"Theorem 11.2(i) shows that a monotonic subsequence of $s_n$ converges to $t = \lim \sup s_n$"
However, Theorem 11.2(i) does not require the subsequence of $s_n$ converging to $t$ to be monotonous, which is something that theorem 11.7 requires.
Theorem 11.7:
Let ($s_n$) be any sequence. There exists a monotonic subsequence whose limit is $\limsup s_n$ and there exists a monotonic subsequence whose limit is $\lim \inf s_n$.
Proof:
If ($s_n$) is not bounded above, then by Theorem 11.2(ii), a monotonic subsequence of ($s_n$) has limit $+\infty = \limsup s_n$. Similarly, if ($s_n$) is not bounded below, a monotonic subsequence has limit $-\infty = \liminf s_n$.
The remaining cases are that ($s_n$) is bounded above or is bounded below. These cases are similar, so we only consider the case that ($s_n$) is bounded above, so that lim sup $s_n$ is finite. Let $t = \limsup s_n$, and consider $\epsilon > 0$. There exists $N_0$ so that
$\sup\{s_n : n > N\} < t + \epsilon$ for $ N \geq N_0$.
In particular, $s_n$ $ < t + \epsilon $ for all $n > N_0$. We now claim $\{n \in \mathbb{N} : t - \epsilon $ < $s_n$ < $t + \epsilon\}$ is infinite.
Otherwise, there exists $N_1 > N_0$ so that $s_n$ $\leq t - \epsilon$ for $n > N_1$. Then $\sup \{s_n : n > N\} \leq t - \epsilon$ for $N \geq N_1$, so that $\limsup s_n < t$, a contradiction. Since (1) holds for each $\epsilon > 0$, Theorem 11.2(i) shows that a monotonic subsequence of $(s_n)$ converges to $t = \limsup s_n$.
Theorem 11.2(i):
If $t$ is in $\mathbb{R}$, then there is a subsequence of $(s_n)$ converging to $t$ if and only if the set $\{n \in \mathbb{N} : |s_n - t| < \epsilon\}$ is infinite for all $\epsilon > 0$.
Theorem 11.2 (ii):
If the sequence $(s_n)$ is unbounded above, it has a subsequence with limit $+\infty$.