Let $(\mathsf{X},\mathcal{X})$ be a measurable space, $\mathcal{F}_b(X)$ be the space of bounded measurable functions, and define the supremum norm as $\Vert f \Vert_{\infty} = \sup\{ f(x) : x \in \mathsf{X}\}$ (absolute value?).
Why is $$ \sup_{(x,x') \in \mathsf{X} \times \mathsf{X}}|f(x) - f(x')| = 2\inf_{c \in \mathbb{R}}\Vert f - c \Vert_{\infty}? $$
The left hand side is the oscillation seminorm, and it seems like it's the peak minus the trough of $f$, while the right hand side seems like it should be $-\infty$ or $0$.
It's only correct if you define the supremum norm with the absolute value: $||f||_\infty := \sup |f(x)|$.
Then, for $b=(\sup(f)+\inf(f))/2$, and $g(x)=f(x)-b$, we have $$\sup(g(x))=\frac{\sup(f(x))-\inf(f(x))}{2} \qquad \text{and}\qquad \inf(g(x))=\frac{\inf(f(x))-\sup(f(x))}{2}.$$
Thus,
\begin{align*} 2||g||_\infty=|\sup(f(x))-\inf(f(x))|=\sup_{(x,x')\in X\times X} |f(x)-f(x')|. \end{align*}
This gives one inequality
\begin{align*} \sup_{(x,x')\in X\times X} |f(x)-f(x')|\geq 2\inf_{c\in \mathbb{R}}||f-c||_\infty. \end{align*}
Now, you can use similar ideas to show equality must hold.