I'm reading the proof from Hildebrand that for some $c_1 > 0$, the Riemann zeta function $\zeta(s)$ has no zero in the region $\sigma > 1-c_1$, $|t| \le 2$. (Here $s = \sigma + it$ per tradition).
After showing that $\zeta$ is zero-free for $\sigma \ge 1$, the author writes:
. . . since $\zeta(s)$ has no zeros in the closed half-plane $\sigma \ge 1$, so $1 / \zeta(s)$ is analytic in this half-plane and therefore bonuded in any compact region contained in this half-plane. In particular, $1/\zeta(s)$ is bounded in the rectangle $1 \le \sigma \le 2$, $|t| \le 2$. By compactness, it follows that $1/\zeta(s)$ remains bounded in any sufficiently small neighborhoood of this rctangle, and, in particular, in a rectangle of the [desired form].
I have two questions about this:
(i) Why is the bolded statement true?
(ii) Where does this argument break down if we replace $\zeta(s)$ by $L(s, \chi)$? After all, $L(s, \chi)$ also has the property of being zero-free in the region $\sigma \ge 1$, but we don't have a similar conclusion for $L(s, \chi)$.