An exercise asked us to prove that Lipschitz-continuity implies uniform continuity. I did it by picking $\delta = \frac{\varepsilon}{L}$ for $L > 0$ and $\delta = \varepsilon$ for $L=0$. The sample solution for the series has almost same result. Here is the solution in German, exercise 1.
I compared my solution with a fellow student and he had a different solution for this exercise, where he defined $\delta = \frac{\varepsilon}{L + 1}$ to avoid the division by zero case. I tried out some examples with his $\delta$, while it does appear to work, a lot of "$\varepsilon$ space" is unused.
- Is his idea correct?
- Is there a rule for how well a $\delta$ needs to fit the $\varepsilon$ area?
- Is it possible to create seemingly correct proofs which are actually wrong like that?
For the case where $L=0$ case it makes most sense to observe that it means that the map is constant. To let $\delta = \frac{\varepsilon}{L+1}$ works for both cases because
$$|f(x)-f(y)| \leq L|x-y| < L \frac{\varepsilon}{L+1} = \frac{L}{L+1} \varepsilon < \varepsilon.$$
When it comes to your concern about "unused $\varepsilon$-space", there is usually no reason to look for "the best" $\delta$ when doing proofs like this. If you were doing an error estimate in a numerical analysis, it would be another matter, but in this case that should not be a concern.
As to question 3., it is definitely possible to create proofs that seems convincing, but actually contains minor or major flaws. I hope this answers your questions.