Boyd & Vandenberghe, question 2.31(d). Stuck on simple problem regarding interior of a dual cone
Arnaud is correct.  The critical piece I was missing is that for any $z$, $z^\top x &gt; 0 \iff z^\top x / \|x\| &gt; 0.$

So pick $z \in S$. Let $x_0 = \text{arg inf } \{ z^\top x \mid x \in \text{cl}(K) \cap \{0\}, \|x\|=1 \}$
and let $c_{x_0} = z^\top x_0 &gt; 0$.  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

$~~~~ z'^\top x_0 \ge c_{x_0} - \gamma &gt; c_{x_0} / 2 &gt; 0.$

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

$~~~~ z'^\top x \ge z'^\top x_0 &gt; 0.$

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

$~~~~ z'^\top x / \|x\| &gt; 0$

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