Here's how I did it:
Suppose $x_n \rightarrow \infty$. Hence, for any $M \in \mathbb{R}$, we can find some $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n \geq M$. Then, for such $N$, we take $N^{th}$ tail of that sequence to have:
inf ($T_n) \geq M$ and sup $(T_n) \geq M$.
Since we can do this for any $M,$ it is implied that
lim inf $x_n$ = lim sup $x_n$ = $\infty$.
The proof for the second part is very similar.
Your proof is correct. If you want to be a little more precise:
Suppose $x_n\to\infty$.
For $n\in\mathbb{N}$, define $$ T_n:=\{x_m:m\geq n\}\\ A_n:=\inf\left(T_n\right) $$
Note that if $m\geq n$, then $T_m\subseteq T_n$ so $$ A_m\geq A_n\tag{1} $$
Let $M\in\mathbb{R}$. Since $x_n\to\infty$, there exists $N\in\mathbb{N}$ such that $x_n\geq M$ if $n\geq N$. By definition of infimum as the greatest lower bound, it follows that $A_{N}\geq M$. By $(1)$, it follows that if $n\geq N$ then $A_n\geq M$.
This shows that $$ \lim_{n\to\infty}A_n=\infty $$
Hence, by definition of limit inferior, $\liminf x_n=\infty$.
Note. If $x_n\to\infty$, then $\{x_n\}$ is not bounded above and $\sup T_n$ is always infinite. So if $\{x_n\}$ is not bounded above, then we usually define $\limsup x_n$ to be $\infty$.
Your proof looks correct. Also, you only need $\liminf x_n=\infty$ in order to conclude $x_n\rightarrow\infty$ (the part with $\limsup$ is superfluous).
Just come back to definitions, when you hesitate.
Let's have $v_n=inf\,\{x_k\,|\,k\ge n\}$ and $w_n=sup\,\{x_k\,|\,k\ge n\}$.
$w_n\ge v_n$ so we only have to proove that $v_n\to+\infty$ and $w_n\to+\infty$ is automatic.
From $(x_n)_n$ divergence we have $\forall M\in\mathbb R\quad v_N\ge M$
Thus since $(v_n)_n$ is increasing we get $\forall n\ge N\quad v_n\ge v_N\ge M$, which is by definition $v_n\to+\infty$.
And since $\underline\lim x_n=\lim v_n$ and $\overline\lim x_n=\lim w_n$ the demonstration is done.