Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).

Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.

Let $A \subseteq C(K)$ be non-empty. Define

$ g(x) = \sup_{f \in A} f(x)$

$ h(x) = \inf_{f \in A} f(x)$

$g$ is upper semicontinuous: $g(x) > a$ iff there exists $f \in A$ such that $f(x) > a$. Similarly $h$ is lower semicontinuous.

Further, $g(x) - h(x) \leq \textrm{diam}(A)$.

Therefore $g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$

But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with

$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$

But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.

But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.

Edit: This proof is wrong, but preserving for posterity.

Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).

Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.

Let $A \subseteq C(K)$ be non-empty. Define

$ g(x) = \sup_{f \in A} f(x)$

$ h(x) = \inf_{f \in A} f(x)$

$g$ is upper semicontinuous: $g(x) > a$ iff there exists $f \in A$ such that $f(x) > a$. Similarly $h$ is lower semicontinuous.

Further, $g(x) - h(x) \leq \textrm{diam}(A)$.

Therefore $g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$

But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with

$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$

But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.

But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.