Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c$ and $|f(b)| = c$. Is there a $\delta (\varepsilon)$ with $\|f\|_{\infty}< \delta (\varepsilon) $ such that we can find Lipschitz function $g$ such that $|g| \geq c,$ $g(a)=f(a),$ $ g(b)=f(b)$ and Lipschitz constant of $f-g$ is less than any $\varepsilon$?

There should be some simple counterexample.

It is possible that $\|f-g\|_{\infty}< \varepsilon$. Indeed, let $\delta= \frac{\varepsilon}{2}$. Then we can take $g$ with $\|g\|_{\infty}< \delta$.

Is it possible if we replace condition $\|f\|_{\infty}< \delta (\varepsilon)$ by: Lipschitz constant of $f$ is less than $\delta (\varepsilon) $?