POTW: Proving $\lim\limits_{x\to \infty} |f(x)| = \infty$

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Here is this week's POTW:

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Prove that if $f:\mathbf{R} \to \mathbf{R}$ is continuous and $\lim\limits_{x\to \infty} f(f(x)) = \infty$, then $\lim\limits_{x\to \infty} |f(x)| = \infty$.
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No one solved this problem. You can read my solution below.

Spoiler

If the conclusion is false, there is an $M > 0$ and a sequence $x_n \to \infty$ such that $|f(x_n)| \le M$ for all $n\in \mathbb{N}$. By the Bolzano - Weierstrass theorem, $\{f(x_n)\}$ has a convergent subsequence $\{f(x_{n_k})\}$. Suppose $f(x_{n_k}) \to L$. Continuity of $f$ implies $f(f(x_{n_k})) \to f(L)$. This contradicts the assumption $\lim\limits_{x\to \infty} f(f(x)) = \infty$.

Edit: Sorry, I've overlooked Opalg's solution, which you can read below!

Spoiler

Proof by contradiction (it involves knowing how to negate existential statements!).

Suppose that the result is false. Then there is a sequence $(x_n)$ such that $\lim\limits_{n\to\infty}x_n = \infty$ but $|f(x_n)|$ does not tend to infinity as $n\to\infty$. This means that there exists $R\in\mathbf{R}$ such that $|f(x_n)|\leqslant R$ infinitely often. So there is a subsequence $x_{n_k}$ (which satisfies $\lim\limits_{k\to\infty}x_{n_k} = \infty$) such that $|f(x_{n_k})| \leqslant R$ for all $k$.

The function $f$ is continuous, so it is bounded on closed bounded subsets of $\mathbf{R}$. Since the interval $[-R,R]$ is closed and bounded, it follows that there exists $M\in\mathbf{R}$ such that $f(x) \leqslant M$ whenever $|x|\leqslant R$.

Putting all that together, it follows that $f(f(x_{n_k})) \leqslant M$ for all $k$. But that contradicts the fact that $\lim\limits_{x\to\infty}f(f(x)) = \infty$.