This comes from one of the exercises in How to Prove It by Daniel Velleman. In this proof I invoke the theorem from the previous exercise: “for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$.” I won’t include that proof here, as I am virtually certain it is valid. Anyway, here is my proof:
Theorem. For any real number $x$, $-|x| \leq x \leq |x|$.
Proof. Let $x$ be an arbitrary real number. We can of course write $x = x$, as this is a tautology. By taking the absolute value of both sides of this equation, it follows that $|x| = |x|$. Given this, by the theorem in the previous problem we can conclude that $-|x| \leq x \leq |x|$. Since $x$ was arbitrary, this is true for all real numbers $x$.
I apologize if this seems basic, but I’m really doubting this proof, and I couldn’t find anything similar in my searches.