Let $M(f,A)=sup\{f(x):x\in A\subseteq[a,b]\}$ and let $m(f,A)=inf\{f(x):x\in A\subseteq[a,b]\}$.
Given that $|f(x_{0})|-|f(y_{0})|\leq |f(x_{0}-f(y_{0})|\leq M(f,S)-m(f,S)$ for $x_0,y_0\in S$, prove that $M(|f|,S)-m(|f|,S)\leq M(f,S)-m(f,S)$.
By the given inequality we see that $M(f,S)-m(f,S)$ is an upper bound of $\{|f(x_{0})|-|f(y_{0})|:x_0,y_0 \in S\}$. I was given a hint providing that $sup\{|f(x_{0})|-|f(y_{0})|:x_0,y_0 \in S\}=M(|f|,S)-m(|f|,S)$.
However, I'm not exactly sure to go about proving it is in fact the supremum of $\{|f(x_{0})|-|f(y_{0})|:x_0,y_0 \in S\}$.
Could anyone clarify on how to prove that $sup\{|f(x_{0})|-|f(y_{0})|:x_0,y_0 \in S\}=M(|f|,S)-m(|f|,S)$?
For every $x$ and $y$ in your set $$f(x) \leq \mathbf{sup} _{x}f(x)=M$$ $$f(y) \geq \mathbf{inf}_{y}f(y)=m \implies -f(y) \leq -\mathbf{inf}_{y}f(y)=-m$$ Add the two up to get
$$f(x)-f(y) \leq M-m.$$ Remember that this holds for all $x$ and $y$, so, in particular if we pick $y$ for $x$ and vice versa: $$f(y)-f(x) \leq M-m.$$ The last two mean $$|f(x)-f(y)| \leq M-m.$$ Finally,
$$|f(x)|-|f(y)| \leq |f(x)-f(y)| \leq M-m\ \ (*) $$ Holds for all real numbers by the triangle inequality.
Now, in (*) forget about the middle term. The left hand side is independent of variables, a real number really. Fix $y$, and take sup with respect to $x$:
$$ \mathbf{sup}_x|f(x)|-|f(y)| \leq M-m \ \ \ (**)$$
Now, (**) holds for all $y$. Take inf with respect to $y$: $$ \mathbf{sup}_x|f(x)|-\mathbf{inf}_y|f(y)| \leq M-m \ \ \ (**)$$ But RHS is just what you wanted. Done. Done.