You want to prove that $\lim\limits_{x\to 1}(x+4) = 5$ using $\epsilon$-$\delta$.
Let $\epsilon\gt 0$. We need to prove that there exists a $\delta\gt 0$ such that
If $0\lt |x-1|\lt \delta$ then $|f(x)-5|\lt \epsilon$.
Now, we want to think a bit: how will the size of $|x-1|$ affect the size of $|f(x)-5|$? Since $f(x)=x+4$, we notice that $|f(x)-5| = |(x+4)-5| = |x-1|$; that is, the size of $|f(x)-5|$ is equal to the size of $|x-1|$. So in order to make sure that $|f(x)-5|\lt \epsilon$, it is enough to require that $|x-1|\lt\epsilon$.
Thus, we can select $\delta=\epsilon$. Then $\delta\gt 0$, and if $0\lt |x-1|\lt\delta$, then it will follow that $|f(x)-5|\lt\epsilon$.
Thus, for all $\epsilon\gt 0$ there exists a $\delta\gt 0$ (namely, $\delta=\epsilon$) with the property that if $0\lt |x-1|\lt \delta$, then $|f(x)-5|\lt \epsilon$. This proves that $\lim\limits_{x\to 1}f(x) = 5$, as desired. $\Box$
That's what you have, only with lots of words thrown in in-between...
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).
Best Answer
The construction of an $\epsilon$-$\delta$ proof is usually exactly opposite its presentation. This should be a good example. Given $\epsilon$, you need to be able to produce some $\delta$ such that for all $x$ with $|x-1|<\delta$, $\frac{2 + 4x}{3} - 2|<\epsilon$.
How do you do this? You (usually) need some formula to produce $\delta$ in terms of $\epsilon$. So start with the expression you need, $$\bigg|\frac{2 + 4x}{3} - 2\bigg|<\epsilon,$$ and start solving for $x$ in terms of $\epsilon$.
I'll let you work out the details here; it's a simple algebraic exercise. At the end, you'll get something like $$\bigg|x - 1\bigg| < \frac{3}{4}\epsilon.$$
Ah-hah! Now you have produced a constraint on $|x-1|$ in terms of $\epsilon$. If I give you $\epsilon$, you can pick $\delta < \frac{3}{4}\epsilon$ and, as you can quickly verify, this $\delta$ will satisfy the $\epsilon$ bound. In fact, it has to --- that's how you made it.
The take-home lesson here, again, is that the way you build the proof is backwards. Start with what you want and backsolve for what you need. Then when you write the proof out, you know how to choose $\delta$, so you can quickly verify that $\lim_{x\to 1}\frac{2 + 4x}{3} = 2.$