Prove that $\lim_{n\to\infty} \sup{x_{n}}=\lim_{n\to\infty} \inf{x_{n}}=L$ IFF the sequence $(x_n)$ converges.

Prove that $\lim_{n\to\infty} \sup{x_{n}}=\lim_{n\to\infty} \inf{x_{n}}=L$ IFF the sequence $(x_n)$ converges.

My attempt: $\Rightarrow$ Since $\inf{x_n}\leq x_{n} \hspace{1mm} \forall n\in\mathbb{N}$ then we have that $$\lim_{n\to\infty} \inf_{k\geq n}{x_k}<\lim_{n\to\infty}x_n$$ Similarly we have that $x_{n}<\sup{x_n}\hspace{1mm} \forall n\in\mathbb{N}$, then we obtain for some $n\in\mathbb{N}$ $$\lim_{n\to\infty} x_n<\lim_{n\to\infty} \sup_{k\geq n}{x_k}$$ Combining inequalities we obtain $$\lim_{n\to\infty} \inf_{k\geq n}{x_k}<\lim_{n\to\infty}x_n<\lim_{n\to\infty} \sup_{k\geq n}x_k$$ Then from the assumption that $\lim_{n\to\infty}\sup{x_{n}}=\lim_{n\to\infty}\inf{x_n}=L$ we obtain that $\lim_{n\to\infty}x_n=L$

$\Leftarrow$ If the sequence converges to some $L\in\mathbb{R}$ then there exists some $N\in\mathbb{N}$ such that $\forall n\geq N$ and $\varepsilon >0\hspace{2mm}$ we have $\left | x_n-L\right |<\varepsilon$ then we obtain $-\varepsilon+L<x_n<\varepsilon+L$. Since $\inf{x_n}<x_n<\sup{x_n}$ we have $-\varepsilon+L<\inf{x_n}\leq \sup{x_n}<\varepsilon+L$. $|\inf{x_n}-L|<\varepsilon$ and $|\sup{x_n}-L|<\varepsilon$, since $\varepsilon$ is arbitrary we conclude that $\lim_{n\to\infty}\sup{x_n}=\lim_{n\to\infty}\inf{x_n}=L$

I stopped reading when I saw the first inequality. We have $\inf_{n \geq k}x_{n} \leq x_{n}$ for all $n \geq k$, and hence $$\liminf_{n \to \infty}x_{n} = \lim_{k \to \infty}\inf_{n \geq k}x_{n} \leq \lim_{n \to \infty}x_{n};$$ the inequality involving $\limsup$ holds likewise.
Though I am not sure what the definitions of limit superior and limit inferior you are using, anyway either by definition or by proof we have the following fact: If $l_{1},l_{2} \in \Bbb{R}$, then $l_{1} = \limsup_{n \to \infty}x_{n}$ iff for every $\varepsilon > 0$ we have $x_{n} < l + \varepsilon$ for large $n$ and $x_{n} > l-\varepsilon$ for infinitely many $n$; moreover, we have $l_{2} = \liminf_{n \to \infty}x_{n}$ iff for every $\varepsilon > 0$ we have $x_{n} < l+\varepsilon$ for infinitely many $n$ and $x_{n} > l-\varepsilon$ for large $n$.
If $l := \limsup_{n \to \infty}x_{n} = \liminf_{n \to \infty}x_{n}$, then for every $\varepsilon > 0$ we have $x_{n} > l - \varepsilon$ and $x_{n} < l + \varepsilon$ for large $n$ by the above result; hence $x_{n} \to l$. Conversely, if $x_{n} \to l$ for some $l \in \Bbb{R}$, then $x_{n} < l+\varepsilon$ and $x_{n} > l-\varepsilon$ for large $n$; again by the above result we have $l = \limsup_{n \to \infty}x_{n} = \liminf_{n \to \infty}x_{n}$.