real analysis - If $lim_{xrightarrowinfty}f(f(x))=infty$, $f:(-infty,infty)rightarrowmathbb{R}$ is continuous, prove $lim_{xrightarrowinfty}|f(x)|=infty$

I'm working on another homework problem that I could use some help with. The question is posed in the title, and given here (exactly):

Problem: Let $f:(-\infty,+\infty)\rightarrow\mathbb{R}$ be continuous and the $\!\!\lim\limits_{x\rightarrow+\infty}\!\!f(f(x))=+\infty$. Prove that $\!\!\lim\limits_{x\rightarrow+\infty}\!\!|f(x)|=+\infty$.

It is a former, prelim problem from earlier in the year, and that was why I was requesting help - please note, I'm finishing up an honors, undergraduate degree in Mathematics, so this is a regular homework problem. As we speak, I'm trying to prove this by contradiction. If I'm correct, we have $\exists\varepsilon>0:\forall M\in\mathbb{N},\exists x\in(-\infty,+\infty):x>M~\text{and}~|f(x)|>\varepsilon$ negating the sufficient part, or we can suppose that $\lim\limits_{x\rightarrow\infty}|f(x)|\neq +\infty$ so that $\lim\limits_{x\rightarrow\infty}|f(x)|=L$. I've been working with both, going back and forth, using the precise definitions overall in order to get a contradiction against one of the two given assumptions using the other. I'm not sure if this is the way to go, and any help is GREATLY appreciated!

Answer

The negation of $\lim\limits_{x\rightarrow\infty}{|f(x)|}=\infty$ is: There exists $M$ such that for every $N\in\mathbb{R}$, there exists $x>N$ such that $|f(x)|

Assuming this negation, we can obtain a sequence $\{x_n\}$ such that $x_n\rightarrow\infty$ and $f(x_n)$ converges to some finite limit $L$ (why can we do this?). We then look at what $\lim\limits_{n\rightarrow\infty}{f(f(x_n))}$ must be.