Boyd & Vandenberghe, question 2.31(d). Stuck on simple problem regarding interior of a dual cone.

Okay, the critical piece I was missing is that for any $z$, $z^\top x > 0 \iff z^\top x / \|x\| > 0.$

So let $x_0 = \text{arg inf } \{ z^\top x \mid x \in \text{cl}(K), \|x\|=1 \}$ and let $c_{x_0} = z^\top x_0 > 0$. We know $x_0$ exists because the set $\text{cl}(K)\cap \{x \mid \|x\| = 1 \}$ is closed and bounded.

Now let $\varepsilon = c_{x_0} / 2$ so that for any $z' = z + \gamma u$ where $\|u\|=1$ and $\gamma \in (0,\varepsilon)$, we have

$$z'^\top x_0 \ge c_{x_0} - \gamma > c_{x_0} / 2 > 0.$$

Furthermore, for any other $x \in \text{cl}(K)\cap \{x \mid \|x\| = 1 \}$ we know

$$z'^\top x \ge z'^\top x_0 > 0.$$

Therefore, for any $x \in \text{cl}(K) \backslash \{0\}$ we have

$$z'^\top x / \|x\| > 0$$

which proves $z'^\top x > 0$ whenever $z' \in D(z,\varepsilon)$.

The other piece of the proof which is to show that $z^\top x = 0$ for $x \in \text{cl}(K) \backslash \{0\}$ only when $z$ is on the boundary of $K^*$. This isn't very difficult.

The question has an additional bug: $K$ should be assumed to be closed. For instance, take $K$ to be the positive orthant together with $0$, i.e., $K=\{x\in\mathbb{R}^{n}:x_{i}>0,i=1,\ldots n\} \cup \{0\}$. Then $K^{\ast}=\mathbb{R}_{+}^{n}=\{x\in \mathbb{R}^{n}:x_{i}\geq0,i=1,\ldots n\}$. Every element $z\in K^{\ast}\backslash \{0\}$ satisfies $z^{T}x>0$ for all $x\in K\backslash \{0\}$, even those at the boundary of $K^{\ast}$.