Prove that if f is a bounded function on a nonempty set S, then |sup f| is less than or equal to sup|f|.
Obviously
$\displaystyle f(x)\leq|f(x)|$ for all x, so it holds for supremum:
$\displaystyle \sup_{x} f(x)\leq \sup_x |f(x)|$
Then we have
$\displaystyle - \sup_x |f(x)| \leq \sup_{x} f(x) \leq \sup_x |f(x)|$
hence
$\displaystyle |\sup_{x} f(x) | \leq \sup_x |f(x)|$