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 $\max \{\vert \max_D f(x) \rvert, \vert \min_D f(x) \rvert \} = \max_D\vert f(x) \vert$