I'll take the definition of $\Vert f\Vert_\infty=\Vert f\Vert$ to be $ \Vert f\Vert= \inf\bigl\{ \alpha>0\mid \mu\bigl( \{ x\ \bigl|\ |f(x)|>\alpha\}\bigr)=0\bigr\}. $ Note that
$\tag{1} \beta>\Vert f\Vert \quad\Longrightarrow\quad\mu\bigl(\{ x \mid |f(x)|>\beta\}\bigr)=0. $
To show $|f(x)|\le\Vert f\Vert$ for almost all $x$:
Let $M=\bigl\{ x \mid |f(x)|>\Vert f\Vert\bigr\}$. For each positive integer $n$, set $M_n=\bigl\{ x \mid |f(x)|>\Vert f\Vert+{1\over n}\bigr\}$. Then by $(1)$ we have $\mu(M_n)=0$ for all $n$. Note that $M=\bigcup_{n=1}^\infty M_n$. Now note that, since a countable union of sets of measure zero has measure zero:
$\mu\bigl(\{x\mid |f(x)|> \Vert f\Vert \bigr) =\mu\Bigl(\bigcup_{n=1}^\infty M_n\Bigr) =0. $ Consequently, $|f(x)|\le \Vert f\Vert $ almost everywhere.
To show that if $A < ||f||_\infty$, then there exists a set $E \in \mathcal{X}$ with $\mu(E) > 0$ such that $|f(x)| > A$ for all $x \in E$:
Suppose that $A<\Vert f\Vert$. By the definition of $\Vert f\Vert$ as the infimum of all $\alpha>0$ such that $\mu\bigl(\{ x \mid |f(x)|>\alpha\}\bigr)=0$, it follows that $\mu\bigl(\{ x \mid |f(x)|>A\}\bigr)\ne0$. So simply take $E=\{ x\mid |f(x)|>A\}$.