You made a mistake in the implication that
$$ -99 < x < 101 \implies -\frac{1}{99} < \frac{1}{x} < \frac{1}{100} $$
When $A, B$ are positive numbers, you have (note the reversal of inequality signs)
$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B}$$
When $A,B$ are both negative numbers, you also have (the same thing)
$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} > \frac{1}{B} $$
But when $A < 0 < B$, all you can say is
$$ A < x < B \implies \frac{1}{A} > \frac{1}{x} \mbox{ or } \frac{1}{B} < \frac{1}{x} $$
in particular, the absolute value
$$ \frac{1}{|x|} > \min( \frac{1}{|A|}, \frac{1}{|B|} )$$
is not bounded above a priori.
To address your query
For some reason, I don't like the fact that we can let delta be something and then later "rectify" it after putting in a form of epsilon
perhaps it would be clearer if the $\epsilon$-$\delta$ statements were written this way (I'll use the case of the limit as an example, but you can substitute similar statements into other definitions):
Given a function $f(x)$ and a point $x_0$, and a value $y$. We say that $\lim_{x\to x_0} f(x) = y$ if there "we can find" a function $\delta(\epsilon)$ (a function $\delta$ that depends on the variable $\epsilon$) such that the expression $|f(x) - y| < \epsilon$ holds true whenever $0 < |x-x_0| < \delta(\epsilon)$.
Rmk: I put "we can find" in quotes because in non-school mathematics (as practised by professional mathematicians), we usually cannot write down an explicit formula for the function $\delta(\epsilon)$. We just can demonstrate that such a function must exist.
In particular, doing an $\epsilon$-$\delta$ proof is like reducing a system of algebraic equations: you want to "solve" for the function $\delta(\epsilon)$ (subject to the above caveat).
You are given that $\lim_{x\rightarrow a}f(x)=L$. That means given any "positive value" there exists "another positive value"(depends on the "positive value") such that
if $0<|x-a|<$"another positive value" then $|f(x)-L|<$"positive value". This is the fact we have.
Now, you need to show that given any $\epsilon>0$ there exists $\delta>0$, such that $0<|x-a|<\delta$ then $|cf(x)-cL|<\epsilon$.
So, first take any $\epsilon>0$. Then $\epsilon/|c|$ is also positive. Then by the fact we have, there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon/|c|$. That means $|cf(x)-cL|<\epsilon$. So, we are done.
Best Answer
There is a thought process here which might feel circular, but the proof itself is not. Keep in mind that I'm allowed to use however silly a thought process I want - the only thing that matters is whether the actual proof produced at the end of the day is valid. If I got my inspiration for what $\delta$ should be by rolling dice, well, that might not be something I should rely on in the future but that doesn't mean that I won't be able to turn that guess into a valid proof.
The general thought process in an $\epsilon$/$\delta$ argument is to start with the conclusion we want and try to "backsolve" for what choice of $\delta$ (in terms of $\epsilon$) would work. This does give a circular taste to the whole experience of discovering and subsequently writing the proof, since we seem to start at the end, go to the beginning, and then go back to the end. But the proof itself consists only of the second half of that development.