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.
No, not even when $\lambda>0$. Consider $A=\lambda I+K$ where $\lambda$ is a small positive number and $K$ is a nonzero skew-symmetric matrix. Then $\langle Ax,x\rangle=\lambda\|x\|^2$ for all vectors $x$. However, when $x$ is chosen such that $y=Kx\ne0$, we have
\begin{aligned}
\langle Ax,y\rangle^2
&=\left(\lambda\langle x,y\rangle+ \langle Kx,y\rangle\right)^2\\
&=\left(\lambda\langle x,y\rangle+\|y\|^2\right)^2\\
&>\lambda^2\|x\|^2\|y\|^2\\
&=\langle Ax,x\rangle\langle Ay,y\rangle
\end{aligned}
when $\lambda$ is sufficiently small.
The inequality in question is true when $A$ is symmetric and $\lambda\ge0$. In this case the condition $\langle Ax,x\rangle\ge\lambda\|x\|^2$ implies that $A$ is positive semidefinite. Therefore $(x,y)\mapsto\langle Ax,y\rangle$ defines a semi-inner product and Cauchy-Schwarz inequality is satisfied.
Alternatively, since the quadratic form $\langle Ax,x\rangle$ depends only on the symmetric part $S=\frac{A+A^T}{2}$ of $A$, when there is some $\lambda\ge0$ such that $\langle Ax,x\rangle\ge\lambda\|x\|^2$ for all $x$, we see that $S$ is positive semidefinite. Therefore, Cauchy-Schwarz inequality gives $\langle Sx,y\rangle^2\le\langle Sx,x\rangle\langle Sy,y\rangle$ and consequently,
$$
\left(\frac{\langle Ax,y\rangle+\langle x,Ay\rangle}{2}\right)^2
\le\langle Ax,x\rangle\langle Ay,y\rangle.
$$
Best Answer
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, $\lvert\langle u, v \rangle\rvert \leqslant \|u\|\|v\|$.