I assume that you are assuming that $\max$ and $\min$ exists or that you are assuming that $f(x)$ is continuous which in-turn guarantees that $\max$ and $\min$ exists, since $D$ is compact.
First note that if we have $a \leq y \leq b$, then $\vert y \vert \leq \max\{\vert a \vert, \vert b \vert\}$, where $a,b,y \in \mathbb{R}$. Hence, $$\min_D f(x) \leq f(x) \leq \max_D f(x) \implies \vert f(x) \vert \leq \max\{\vert \max_D f(x) \vert, \vert \min_D f(x) \vert\}, \, \forall x$$ Hence, we have that $$\max_D \vert f(x) \vert \leq \max\{\vert \max_D f(x) \vert, \vert \min_D f(x) \vert\}$$ Now we need to prove the inequality the other way around. Note that we have $$\vert f(x) \vert \leq \max_D \vert f(x) \vert$$ This implies $$-\max_D \vert f(x) \vert \leq f(x) \leq \max_D\vert f(x) \vert$$ This implies $$-\max_D \vert f(x) \vert \leq \max_D f(x) \leq \max_D\vert f(x) \vert$$ $$-\max_D \vert f(x) \vert \leq \min_D f(x) \leq \max_D\vert f(x) \vert$$ Hence, we have that $$\vert \max_D f(x) \rvert \leq \max_D\vert f(x) \vert$$ $$\vert \min_D f(x) \rvert \leq \max_D\vert f(x) \vert$$ The above two can be put together as $$\max \{\vert \max_D f(x) \rvert, \vert \min_D f(x) \rvert \} \leq \max_D\vert f(x) \vert$$ Hence, we can now conclude that $$\max \{\vert \max_D f(x) \rvert, \vert \min_D f(x) \rvert \} = \max_D\vert f(x) \vert$$