I assumed that we work in a real inner product space, otherwise of course we have to put the modulus.
The inequality $\langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$ is also true, but doesn't give any information if $\langle u,v\rangle\leq 0$, since in this case it's true, and just the trivial fact that a non-negative number is greater than a non-positive one. What is not trivial is that $\lVert u\rVert\lVert v\rVert$ is greater than the absolute value. But in fact the assertions
$$\forall u,v \quad \langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$$
and
$$\forall u,v\quad |\langle u,v\rangle|\leq \lVert u\rVert\lVert v\rVert$$
are equivalent. Indeed, the second implies the first, and consider successively $u$ and $-u$ in the first to get the second one.
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
The given argument in the question is not much of a proof, since the "hard" part is to prove that for unit $z$ we have $|<x,z>| \le ||x||$. It certainly admits the interpretation in terms of projection given by cheepychappy, however, this is nothing less than the interpretation of the Cauchy-Schwarz inequality itself.
The proof i recommend is as follows. Consider $x,y$ any vectors in a complex vector space. Then for any $r>0$ and any real $\theta$ expand the inequality $<x+re^{i\theta}y,x+re^{i\theta}y> \ge 0$ to obtain a condition on the roots of a polynomial of second degree in the variable $r$. Express this condition in terms of the coefficients of the polynomial (which will include $||x||, ||y||, |<x,y>|$). What you will get will be the C-S inequality.