"Worst-case" just means to find an upper bound.
Do you understand how he gets the estimate that for all $n\geq N_1$, we have $|b_n|>\frac{|b|}{2}$? If yes, then you just need to remember that when taking reciprocals, the direction of inequalities "flip". Therefore, $\frac{1}{|b_n|}<\frac{2}{|b|}$. We can now multiply both sides of this inequality by the positive number $\frac{1}{|b|}$. Multiplying by positive numbers doesn't change the inequality sign, so $\frac{1}{|b|}\cdot \frac{1}{|b_n|}<\frac{1}{|b|}\cdot \frac{2}{|b|}$, or rewriting this in the usual way, $\frac{1}{|b_n||b|}<\frac{2}{|b|^2}$.
The idea of the proof is that our assumption tells us $b_n\to b$, meaning we only have good control of terms involving $|b_n-b|$ (this is essentially what the definition of convergence is). Now, you should ALWAYS keep in mind the following heuristic principle small times manageable = small.
So, now when we want to prove convergence for the reciprocal, we have to somehow get terms like $|b_n-b|$ to appear, so that we can invoke our hypothesis. In this case, we can do so simply with some algebra. For all $n\geq N_1$, we have $|b_n|>\frac{|b|}{2}>0$, so $b_n\neq 0$, so it actually makes sense to take its reciprocal. Now, simple algebra shows us
\begin{align}
\left|\frac{1}{b_n}-\frac{1}{b}\right|&=|b_n-b|\cdot \frac{1}{|b_n||b|}
\end{align}
The first term $|b_n-b|$ is what we have very good control over; I can make it as small as I want by making $n$ large enough. The second term on the other hand is not necessarily a small quantity. But, that doesn't matter because I have the manageable estimate $\frac{1}{|b_n||b|}<\frac{2}{|b|^2}$. Now is where we invoke the principle that small times manageable is still small.
This is what happens in the proof by saying that given $\epsilon>0$, there is some $N_2$ such that if $n\geq N_2$ then $|b_n-b|<\frac{\epsilon}{2/|b|^2}=\frac{\epsilon |b|^2}{2}$. (This is saying $|b_n-b|$ is small provided $n\geq N_2$). Therefore, by combining the two estimates, we get that for all $n\geq \max(N_1,N_2)$,
\begin{align}
\left|\frac{1}{b_n}-\frac{1}{b}\right|&=|b_n-b|\cdot \frac{1}{|b_n||b|}<\frac{\epsilon |b|^2}{2}\cdot \frac{2}{|b|^2}=\epsilon.
\end{align}
You mentioned that you understood all the laws for converging sequences. So, I suggest you look over the proof for products, and try to identify where the above principle of "small times manageable= small" was used. Perhaps this will help you understand better the proof idea for quotients.
Finally, I should mention that there's actually no need to get a "perfect $\epsilon$" at the end. In other words, the following is also an equivalent condition for convergence (each $\alpha_n,\alpha$ is assumed to be a real/complex number):
A sequence $(\alpha_n)_{n=1}^{\infty}$ converges to $\alpha$ if and only if there is a $K\geq 0$ such that for all $\epsilon>0$, there is some $N\in\Bbb{N}$, such that for all $n\in\Bbb{N}$, if $n\geq N$, then $|\alpha_n-\alpha|\leq K\epsilon$.
I leave it to you to figure out the equivalence.
