Here is the link for the proof:
Proving that $\|f\|_{\infty} = \|f\|_{\max}.$
And here is the proof:
For a continuous function $f : [a,b] \to \mathbb{R}$, we can let $\varepsilon > 0$ be arbitrary and then note that $$U = \{x \in [a,b] \mid |f(x)| > \|f\|_{\max} - \varepsilon\}$$ is an open subset of $[a,b]$ (as a subspace of $\mathbb{R}$). Since $f$ is continuous, $$f(c) = \|f\|_{\max} > \|f\|_{\max}-\varepsilon$$ for some $c \in [a,b]$, $U \ni c$ is nonempty, so $m(U) \neq 0$. Therefore $\|f\|_\infty \geq \|f\|_{\max}-\varepsilon$. Now taking $\varepsilon \to 0^+$ yields $\|f\|_\infty \geq \|f\|_{\max}$.
My question is:
1-I am not sure why he took $f(c) = \|f\|_\infty$ ?
2- Also, it is not so clear for me the general idea of the proof of this direction. Could anyone explain this for me please?
1-They didn't. They took $f(c)=\|f\|_{\max}$ to show $U$ is nonempty. If $U$ was empty, the argument wouldn't work since we wouldn't know there was some $x$ such that $f(x)>\|f\|_{\max}-\varepsilon$ and therefore wouldn't be able to argue
$$(\|f\|_\infty \ge f(c)) \text{ and } (f(c)>\|f\|_{\max}-\varepsilon) \text{ implies } (\|f\|_\infty \ge \|f\|_{\max}-\varepsilon).$$
2- The general idea happens frequently in analysis. If you want to show $a\ge b,$ but it's too hard for whatever reason to do it directly, you can try to show that $a\ge b-\varepsilon$ for all $\varepsilon>0,$ which is what happened here. The next part is figuring out how to write that properly. For example, defining that set $U$ and showing it's nonempty.