How can we show that
$$\left| \inf_S f(x) - \inf_S g(x)\right| \le \sup_S \left| f(x) - g(x)\right|\,?$$
I started with $\left| \sup_S f -g\right|\le \sup_S \left| f- g\right|$. And $\left| \sup_S f-g\right| = \left| \sup_S f -\inf_S g\right|$. But why is it true that
$$\left| \inf f - \inf g\right| \le \left|\sup f - \inf g\right|\,?$$
For an $\epsilon>0$, find some $x\in S$ such that $\inf g+\epsilon>g(x)$, then \begin{align*} \inf f-\inf g-\epsilon&<\inf f-g(x)\\ &\leq f(x)-g(x)\\ &\leq|f(x)-g(x)|\\ &\leq\sup|f-g|, \end{align*} since this is true for all $\epsilon>0$, then $\inf f-\inf g\leq\sup|f-g|$. Swiping the roles of $f$ and $g$, one has $\inf g-\inf f\leq\sup|g-f|=\sup|f-g|$, so we conclude that $|\inf f-\inf g|\leq\sup|f-g|$.