I am doing Exercise 2.3.2 in Tom Lindstrom's real analysis book Spaces. It reads
Show that the function $f:\mathbb{Q} \to \mathbb{Q}$ defined by $f(x) = x^3 - 6x$ is continuous at all $x \in \mathbb{Q}$, but that it does not have a minimum in $[0,2]_{\mathbb{Q}}$. Compare to the Extreme Value Theorem. (We define $[a,b]_{\mathbb{Q}} := \{x \in \mathbb{Q} : 0 \leq x \leq 2\}$.)
We know that $f$ is continuous on all of $\mathbb{R}$ since it is a polynomial, so $f$ is continuous on all of $\mathbb{Q}$. I am having trouble with the next part though.
By looking at the function on $\mathbb{R}$ and using the Closed Interval Method (sort of cheating here to gain some insight), we can check that the minimum of the function on the interval $[0, 2]$ in fact occurs at $\sqrt{2}$ which of course is not in $\mathbb{Q}$, so we know that the function won't have a minimum on $[0, 2]_{\mathbb{Q}}$ since we may get infinitely close to $\sqrt{2}$ in $\mathbb{Q}$. Now, I am trying to prove it rigorously using the tools I have up to this point in the book (sequences, continuity, supremum, infimum, etc.) without using differentiability.
I was originally trying to go by contradiction. Suppose that $f$ achieves its maximum on $[0, 2]_{\mathbb{Q}}$. Then, there exists $m \in [0, 2]_{\mathbb{Q}}$ such that $f(m) \leq f(x)$ for any $x \in [0, 2]_{\mathbb{Q}}$. Then, letting $S := \{f(x) : x \in [0, 2]_{\mathbb{Q}}\}$, we have $f(m) = \inf S$. Then, I was hoping to contradict that this is the infimum possibly using the $\varepsilon$ criteria (not sure how this would use the fact that $m$ is rational though).
The other option would be to proceed directly, showing that for any $m \in [0, 2]_{\mathbb{Q}}$, there exists $n \in [0, 2]_{\mathbb{Q}}$ such that $f(n) \leq f(m)$ (perhaps using the Archimedean property).
Any help would be greatly appreciated. Also, my apologies if anything looks weird in the text. It was rendering oddly on my screen, but I believe everything is typed correctly.