Understanding Rudin Theorem 3.3 (d) For reference, here is the full Theorem 3.3 (d) from Rudin.

Suppose $\{s_n\}$, $\{t_n\}$ are complex sequences, and $\lim\limits_{n \to \infty} s_n = s$, $\lim\limits_{n \to \infty} t_n = t$. Then


(a) $\lim\limits_{n \to \infty} \left(s_n + t_n\right) = s + t$;


(b) $\lim\limits_{n \to \infty} cs_n = cs$,  $\lim\limits_{n \to \infty} \left(c + s_n\right) = c + s$, for any number $c$;


(c) $\lim\limits_{n \to \infty} s_n t_n = st$;


(d) $\lim\limits_{n \to \infty} \frac{1}{s_n} = \frac{1}{s}$, provided $s_n \neq 0$ ($n = 1, 2, 3, \ldots$), and $s \neq 0$.

There's one key step in the proof of (d) that I don't understand. First, Rudin uses convergence of $s_n$ to find an $m \in \mathbb{N}$ so that for all $n \geq m$, we have $|s_n - s| < \frac{1}{2} |s|$. He then asserts that for $n \geq m$, we have
$$ 
|s_n| > \frac{1}{2} |s|.
$$
This is a very important step, but I cannot follow it. I've tried contradiction and the triangle inequality, but I can't get the inequality signs to line up. For example, I tried (for $n \geq m$),
$$ 
|s_n| = |(s_n - s) + s| \leq |s_n - s| + |s| < \frac{1}{2} |s| + |s| = \frac{3}{2} |s|.
$$
It seems as though I've bounded $|s_n|$ "in the opposite direction." If I knew $s$ were positive, expanding the absolute values might work, but we only know it's non-zero.
The rest of the proof looks pretty straightforward to me, with one slight doubt. He asserts the existence of an $N$ (I don't know why he requires $N > m$ when he could just take the maximum of $N$ and $m$; does this make a difference?) so that $n \geq N$ implies $|s_n - s| < \frac{1}{2} |s|^2 \epsilon$. As $|s_n| > \frac{1}{2} |s|$, we have $\frac{1}{|s_n|} < \frac{2}{|s|}$. We then have:
\begin{align*}
\left \lvert \frac{1}{s_n} - \frac{1}{s} \right \rvert & = \left \lvert \frac{s - s_n}{s_n \cdot s} \right \rvert \\
& = \frac{|s - s_n|}{|s||s_n|} \\
& < \frac{2|s - s_n|}{|s|^2} \\
& < \frac{2}{|s|^2} \cdot \frac{1}{2} |s|^2 \epsilon \\
& = \epsilon
\end{align*}
I would appreciate some feedback on the above and some help with those two questions (how Rudin deduces $|s_n| > \frac{1}{2} |s|$ and why he takes $N > m$ instead of $\max(N,m)$.)
 A: === second (more obvious answer) ====
Or we could talk it out:
$|a - b| < k$ means $a$ and $b$ are within $k$ of each other.  Which means $a-k < b < a+ k$.
So $s -\frac 12|s| < s_n < s+\frac 12|s|$ and so:
$\begin{cases} \frac 32 s\\0< \frac 12 s\end{cases} < s_n < \begin{cases}\frac 12s< 0&\text{if }s<0\text{ so }|s|=-s\\\frac 32s &\text{if }s> 0\text{ so }|s|=s\end{cases}$.
(In hindsight, I think Rudin was thinking this argument was obvious.... And it is obvious... in hindsight.)
(Let's face it... Rudin was just smarter than us.....)
==== first answer====
When in doubt do cases or broad strokes or.... whatever
$|s_n - s| < \frac 12 |s|$
$-\frac 12|s| < s_n -s < \frac 12 |s|$.
If $s$ is positive then $s = |s|$ and
$-\frac 12|s| < s_n - |s| < \frac 12|s|$ and
$0 < \frac 12|s| < s_n < \frac 32|s|$ and $s_n > 0$ and $|s_n| = s_n >\frac 12|s|$.
If on the on the other hand $s < 0$ then $|s| =- s$ and so
$-\frac 12|s| < s_n + |s| < \frac 12|s|$ and so
$-\frac 32|s| < s_n < -\frac 12|s| < 0$.
So $s_n < 0$ and $|s_n| = -s_n$ and $\frac 12|s| < -s_n = |s_n|$.
Either way $\frac 12|s| < |s_n| < \frac 32|s|$ and $s_n$ and $s$ are the same sign.
...  There is almost certainly a slicker more direct way with triangle inequality on substituting $t = s_n -s$ so $t+s = s_n$ or something like that but nothing wrong with hitting something with a hammer until it falls apart....
We can do a clunkier way with more hammer strokes but easier calculations by considering the four cases $s_n,s$ positive or negative (as $s_n \to s$ and $s\ne 0$ we can assume $n$ is large enough to assure $s_n$ is the same sign as $s$) and whether $s_n$ is less or greater or equal to $s$...
i.e.
Case 1:  $0 < s < s_n$ so $s_n > s > \frac 12 s > 0$ and $0 < \frac 12|s| < |s_n|$.
Case 2: $0 < s_n \le s$ so $|s_n-s| < \frac 12|s| \implies s-s_n < \frac 12 s\implies \frac 12s < s_n \implies \frac 12|s| < |s_n|$
Case 3: $s_n < s < 0$ so $s_n < s < \frac 12 s <0$ so $0 < -\frac 12s < -s_n$ so $0< \frac 12|s| < |s_n|$
and Case 4:  $s\le s_n < s < 0$ so $|s_n - s| < \frac 12s \implies 0 \le s_n - s < -\frac 12 s \implies s_n < \frac 12 s < 0\implies 0 < \frac 12|s| < |s_n|$.
... but that was pretty tedious and didn't make us look like slick rock stars.
A: By the triangle inequality, $|s|\le |s_n| + |s-s_n|$, or equivalently, $|s-s_n|\ge |s|-|s_n|$. For $n\ge m$ then, $|s-s_n|> \frac{|s|}{2}$.
As to your second question, there is no special reason to take $N \ge m$ or $N> m$, or to let $N'$ be such that $|s_n-s|<\frac12|s|^2\epsilon$ for $n\ge N'$, and then take $N=\max(N',m)$. The goal is simply to ensure there is some constant (which in this case is $N$ because he takes $N > m$) so that for $n$ bigger than that constant, we can ensure both inequalities we want. Taking $N > m$ is a particularly succinct way to do just that.
