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This is probably very obvious, I was looking at this.

It looks so much like a Cauchy-Schwarz though, and I would say it is very obvious from the definition if it wasn't for the condition that $\|x\| \le 1$ in the definition:

$$\| z \| := \sup_x\{ z^T x : \|x\| \le 1\}.$$

How is that restriction still ok?

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In a word, scaling. – Robert Israel Sep 24 '12 at 2:56

Using Robert Israel's hint, let $y = \dfrac{x}{\|x\|}$. Note that $\|y\| = 1$ and $x = \|x\|y$, so we have

$$z^Tx = z^T(\|x\|y) = \|x\|(z^Ty) \leq \|x\|\sup_y\{z^Ty \mid \|y\| \leq 1\} = \|x\|\|z\|_*.$$

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