Is the following argument correct?
Propostition. Suppose $\{x_n\}$ is bounded sequence, $a_n:=\sup\{x_k:k\ge n\}$ as before. Suppose that for some $l\in\mathbf{N}$, $a_l\not\in\{x_k:k\ge l\}$. Then show that $a_j = a_l$ for all $j\ge l$, and hence $\lim\sup x_n = a_l$.
Proof. Let $j\geq l$ and consider that $\{x_k:k\ge l\} = \{x_l,x_{l+1},\dots,x_{j-1}\}\cup\{x_k:k\ge j\}$, thus $\sup\{x_k:k\ge l\} = \max\{\beta,\sup\{x_k:k\ge j\}\}$, where $\beta = \sup\{x_l,x_{l+1},\dots,x_{j-1}\}$, then given our definition of $\{a_n\}$ we have $a_l = \max\{\beta,a_j\}$.
Now evidently $\beta\in\{x_l,x_{l+1},\dots,x_{j-1}\}$, but from hypothesis $a_l\not\in\{x_k:k\ge l\}$, thus $a_l\neq \beta$, thus $\max\{\beta,a_j\}\neq\beta$, consequently $a_l=a_j$, given that our choice of $j\ge l$ was arbitrary it follows that $a_j=a_l,\forall j\ge l$.
$\blacksquare$