My attempt: Let $v$ be the $\lim(\sup(x_n))$.Then according to the definition , There will be infinite number of elements that will lie in the interval $(v-e,v]$where $e>0$.
$v-2e<x_n \le v$ for all $n \ge N$.
Let $v_1$ be the $\lim(\inf(x_n))$.Then according to the definition , There will be infinite number of elements that will lie in the interval $[v_1,v_1+2e)$where $e>0$.
$v_1 \le x_n < v_1+2e$ for all $n \ge N_1$.
Let $K=\max\{N,N_1\}$
AS $v=v_1$.
$2v-2e < 2x_n < 2v + 2e$.
$v-e < x_n < v+e .$ for all $ n \ge K$.
Hence the sequence converges to v . I have written the proof based on my intuition. Is it rigorous and also is it correct?
Suggestion by Copper.hat .
$(x_n)$ is bounded.
$a_n:= \inf \{x_k: k \ge n \}$;
$b_n:= \sup \{x_k: k\ge n \}$;
$a_n \le b_n$;
$\lim_{n} a_n=a=\lim_{n} b_n$;
$a_n =\inf \{x_k|k\ge n \} \le x_n \le$ $\sup \{x_k| k\ge n \}=b_n$;
$a=\lim_{n \rightarrow \infty}a_n \le \lim_{n \rightarrow \infty}x_n \le \lim_{n \rightarrow \infty} b_n =a.$
Hence $\lim_{n \rightarrow \infty}x_n = a$