Let L be the set of limit points of the sequence $x_n$. Just assume that L is closed so that sup L is also a limit point and it is the largest limit point.
Hence, there exists a subsequence $x_{n_m}$ such that $\lim_m x_{n_m} = \sup L$. For every $\epsilon > 0$, there exists an $M$ such that for all $m > M$, we have $|x_{n_m} - \sup L| < \epsilon$. Let $y_M = \sup \{x_{n_m} : m > M \}$. Then, it is clear that $|y_M - \sup L| < \epsilon$. In fact, $|y_p - \sup L| < \epsilon$ for all $p > M, p \in \mathbb{N}$ by the way we have constructed $y_p$. Hence,
$\lim_p y_p = \sup L$ or
$\lim_p \sup \{x_{n_m} : m > p \} = \sup L$.
Since $\sup \{x_n : n > p \} \geq \sup\{x_{n_m} : m > p \}$ $\implies $
$\lim_p \sup \{x_n : n > p \} \geq \lim_p \sup \{x_{n_m} : m > p \}$ $\implies $
$\lim_p \sup \{x_n : n > p \} \geq \sup L$ (ineq 1)
(??) Since $\inf_p \sup \{x_n : n > p \} \leq \sup\{x_{n_m} : m > p \}$ $\implies $
In the above inequality, the LHS is independent of $p$ so taking $\lim_p$ on both sides, we get:
$\inf_p \sup \{x_n : n > p \} \leq \lim_p \sup\{x_{n_m} : m > p \}$ $\implies $
$\inf_p \sup \{x_n : n > p \} \leq \sup L$
But $\inf_p \sup \{x_n : n > p \} = \lim_p \sup \{x_n : n > p \} $ $\implies $
$\lim_p \sup \{x_n : n > p \} \leq \sup L$ (ineq 2)
Therefore, by (ineq 1) and (ineq 2)
$\lim_p \sup \{x_n : n > p \} = \sup L$.
Can someone tell my if (??) is valid? If not, what is the way to show that lim sup $(x_n)$ is sup L? I was only able to show lim sup of a subsequence is equal to sup L?
I am constructing my own proof so I need help in my own working.