Is the following argument correct?
Suppose $\{x_n\}$ is bounded sequence, and $\epsilon>0$ is given. Prove that there exists a $M$ such that for all $k\ge M$ we have $$ x_k-\Bigl(\limsup_{n\to\infty} x_n\Bigr)<\epsilon \qquad\text{and}\qquad \Bigl(\liminf_{n\to\infty} x_n\Bigr)-x_k<\epsilon $$
Proof. We prove the first claim and leave the second as an exercise for the reader.
Define $a_n:=\sup\{x_k:k\ge n\}$, then $x_n\leq a_n$, $\forall n\in\mathbf{N}$, since $\{a_n\}$ is monotonically decreasing and bounded below it converges to $\inf\{a_n:n\in\mathbf{N}\} = \limsup_{n\to\infty} x_n = L$, thus $a_n\geq L$, $\forall n\in\mathbf{N}$ and $a_k-L<\epsilon,\forall k\ge M$ for some $M\in\mathbf{N}$, but then $x_k-L\leq a_k-L<\epsilon,\forall k\ge M$, proving the claim in question.$\qquad\blacksquare$