So, I've been going through Spivak's Calculus (Chapter 5, Limits). I am currently stuck on the proof for Theorem 2 (p. 103) illustrating the limit laws (Sum Rule, Product Rule). I do not quite understand how the values for $\delta$ are determined in the lemma as such:

Lemma (2): If

$$|x - x\_0| < \min \left(1, \frac{\varepsilon}{2(|y\_0| + 1)}\right) \\
|y - y\_0| < \frac{\varepsilon}{2(|x\_0| + 1},
$$

then

$$|xy - x\_0y\_0| < \varepsilon$$

This is followed by the proof:

(2) Since $|x-x\_0| < 1$ we have

$$|x| - |x\_0| \le |x - x\_0| < 1, $$

so that
$$ |x| < 1 + |x\_0|$$

thus
$$ 
|xy - x\_0y\_0| = |x(y-y\_0) + y\_0(x-x\_0)| \\
\qquad \qquad \qquad \ \ \le |x| \cdot |y-y\_0| + |y\_0| \cdot |x-x\_0| \\
\qquad \qquad \qquad \qquad \qquad \qquad \ \ \lt (1 + |x\_0|) \cdot \frac{\varepsilon}{2(|x\_0| + 1)} + |y\_0| \cdot \frac{\varepsilon}{2(|y\_0| + 1)} \\
\lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \ \ \
$$

Two things I mainly do not understand:

1. How is the term $\frac{\varepsilon}{2(|x\_0| + 1)}$ derived? (there was a section earlier on deriving a similar term, but that was for an example on resolving $\lim\_{x \to a} x^2.$ and obtaining $\frac{\varepsilon}{2(|a| + 1)}$ in the process).
2. Why is it that the term $|y\_0| \cdot \frac{\varepsilon}{2(|y\_0| + 1)}$ equates to $\frac{\varepsilon}{2}$?