It is not entirely clear what you mean by 'following Tao's argument'.
Suppose $|\langle v,w\rangle| = \|v\| \|w\|$.
Let $w=\alpha v +h$, with $h \bot v$. Then $\|w\|^2 = |\alpha|^2 \|v\|^2 + \|h\|^2$ and $|\langle v,w\rangle| = |\alpha| \|v\|^2$.
Then the first squared line gives:
$|\alpha|^2 \|v\|^4 = \|v\|^2 (|\alpha|^2 \|v\|^2 + \|h\|^2)$.
Hence either $v=0$ and we have $v = 0 \cdot w$, or $h=0$ in which case $w = \alpha v$.
Alternative approach:
If $w=0$ or $v=0$ the result is true so suppose both are not zero.
Suppose $|\langle v,w\rangle| = \|v\| \|w\|$. Replacing $v$ by $\theta v$, with $|\theta| = 1$ does not change the formula, so
we can assume that
$\langle v,w\rangle = \|v\| \|w\|$.
Note that replacing $(v,w)$ by $(tv, {1 \over t} w)$, with $t>0$ does not change the formula.
Now note that $(t\|v\|-{1 \over t}\|w\|)^2 = \|tv-{1 \over t}w\|^2$.
Now choose $t=\sqrt{\|w\| \over \|v\|}$ to get
$w = {\|w\| \over \|v\|} v$.
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, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.
Best Answer
For some reason, people teaching Linear Algebra forgot to connect inner product with the ordinary dot product in $\mathbb{R}^{3}$ or $\mathbb{R}^{n}$. If you have a vector $u$ and a line through the origin with direction vector $v$, then how would you find the point on that line closest to $u$? Answer: orthogonal projection. The same holds in infinite dimensional inner-product spaces, and even for complex spaces. You want to find a scalar $\alpha$ such that $(u-\alpha v)\perp v$, which gives $$ (u,v)-\alpha(v,v) = 0,\\ \alpha = \frac{(u,v)}{(v,v)} $$ Now you can decompose into a right triangle where $u$ is the hypotenuse, $\alpha v$ is the leg along the line with direction vector $v$ and $u-\alpha v$ is the other leg. Explicitly, you have the following orthogonal decomposition: $$ u = \alpha v + (u-\alpha v),\\ (\alpha v,u-\alpha v) = 0. $$ By the Pythagorean Theorem (which is a direct computation using inner product axioms): $$ \|u\|^{2} =\|\alpha v\|^{2}+\|(u-\alpha v)\|^{2}. $$ The Cauchy-Schwarz inequality is exactly the following $$ \|\alpha v\|^{2} \le \|u\|^{2} \\ \mbox{ with equality iff } \|u-\alpha v\| = 0. $$ Write this out: $$ |\alpha|^{2}\|v\|^{2} \le \|u\|^{2} \\ \frac{|(u,v)|^{2}}{(v,v)^{2}}\|v\|^{2} \le (u,u) \\ |(u,v)|^{2} \le (u,u)(v,v). $$