prove $(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$ prove $(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$.
$|f(x)|\geq f(x)$
$\sup{|f(x)|}\geq\sup{f(x)}$
$(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$ when $\sup{f(x)}\geq0$
...
I don't know how to prove the other half where $\sup{f(x)}<0$.
Any ideas or new approaches? 

For any two functions $f$ and $g$, we have 
$$\sup f(x)+\sup g(x)\ge\sup(f(x)+g(x))$$
We also have $|f(x)|+f(x)\ge0$ for all $x$.  Therefore
$$\sup|f(x)|+\sup f(x)\ge\sup(|f(x)|+f(x))\ge0$$
Using the inequality $\sup|f(x)|\ge\sup f(x)$, which the OP proved, it follows that
$$(\sup|f(x)|)^2-(\sup f(x))^2=(\sup|f(x)|-\sup f(x))(\sup|f(x)|+\sup f(x))\ge0$$
(Remark:  The OP was rightfully concerned that $a\gt b$ does not automatically imply $a^2\gt b^2$ if $b\lt0$.)
