$a_n=\sup \{x_k:k\ge n\}$. $\{x_k:k\ge n+1\}$ is a subset of $\{x_k:k\ge n\}$. Then, $a_n$ is an upper-bound for $\{x_k:k\ge n+1\}$. This implies $\sup\{x_k:k\ge n+1\}=a_{n+1} \le a_n$. Therefore, $a_n$ is monotone decreasing. Let $\limsup x_n =x$. Then, $\lim a_n = x$, which implies that $a_n \ge x$.
Could you check this proof is okay? Thank you in advance.
Edit: I forgot to mention that $x_n$ is bounded.
$a_n$ is monotonically decreasing.
However, we have to be careful that $x$ might be $-\infty$.
$a_n$ is not bounded below if $x_n \to -\infty$.
If $x_n$ is bounded, then $a_n$ is bounded.