If $\|a\| = \|b\| = 0$, then
\begin{align*}
0 & \leqslant \|a + b\|^2 =
\|a\|^2 + \|b\|^2 + 2\langle a, b \rangle =
+2\langle a, b \rangle, \\
0 & \leqslant \|a - b\|^2 =
\|a\|^2 + \|b\|^2 - 2\langle a, b \rangle = -2\langle a, b \rangle,
\end{align*}
therefore $\lvert\langle a, b \rangle\rvert = 0 \leqslant \|a\|\|b\|$.
Suppose, on the other hand, that $\|b\| > 0$. Then a fairly standard argument applies. Define $\lambda = \langle a, b \rangle/\|b\|^2$. Then
$\langle a - \lambda b, b \rangle = \langle a, b \rangle - \lambda \|b\|^2 = 0$, therefore
\begin{align*}
0 & \leqslant \langle a - \lambda b, a - \lambda b \rangle \\
& = \langle a, a - \lambda b \rangle - \lambda\langle b, a - \lambda b \rangle \\
& = \langle a, a - \lambda b \rangle - \lambda\langle a - \lambda b, b \rangle \\
& = \langle a, a - \lambda b \rangle \\
& = \|a\|^2 - \lambda \langle a, b \rangle,
\end{align*}
therefore
$$
\langle a, b \rangle^2 = \lambda\|b\|^2\langle a, b \rangle \leqslant \|a\|^2\|b\|^2,
$$
therefore
$$
\lvert\langle a, b \rangle\rvert \leqslant \|a\|\|b\|.
$$
The argument is similar when $\|a\| > 0$. So the Cauchy-Schwarz inequality holds in all cases.
Addendum
It appears (see my series of shame-faced comments below for details)
that this argument is merely an obfuscation of what is surely the
most "standard" of all proofs of the Cauchy-Schwarz inequality. It is
the one that is essentially due to Schwarz himself, and he had good
reasons for using it, quite probably including the fact that it makes
no use of the postulate that $\langle x, x \rangle = 0 \implies x = 0$!
In a modern abstract formulation, it goes as follows (assuming, of course,
that I haven't messed it up again). For all real $\lambda$, we have
$\|u\|^2 - 2\lambda\langle u, v \rangle + \lambda^2\|v\|^2 =
\|u - \lambda v\|^2 \geqslant 0$.
Therefore, the discriminant of this quadratic function of $\lambda$
must be $\leqslant 0$. That is, $\langle u, v \rangle^2 \leqslant \|u\|^2\|v\|^2$; equivalently, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.
The identity given by Rudin is $\|fg\|_1 \leq \|f\|_2 \|g\|_2$ which is equivalent to Cauchy's inequality:
$\implies\quad$
$
|\langle f, g \rangle|
= \left| \int f \bar{g} \right|
\leq \int |f \bar{g}|
= \int |fg|
= \|fg\|_1
\leq \|f\|_2 \|g\|_2.
$
$\impliedby\quad$
$
\|fg\|_1
= \int |fg|
= \int |f||g|
= \langle |f|, |g| \rangle
\leq \| |f| \|_2 \, \| |g| \|_2
= \|f\|_2 \|g\|_2.
$
Best Answer
Those types of norms are called submultiplicative. Not every norm is submultiplicative, but every operator norm is.