Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!

This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega_1$, which I know relatively little about.

Let $V$ be a normed space. For $A \subseteq V$, define $r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$. Define a *bad sequence* in $V$ to be a sequence $\{ v_\alpha : \alpha < \omega_1 \}$ with the properties that:

$\forall \beta, r(\{ v_\alpha : \alpha < \beta \}) \leq 1$

$\inf_\beta r(\{ v_\alpha : \alpha \geq \beta \}) > 1$

An example of a space with a bad sequence is $c_0(\omega_1)$ (the set of all bounded real-valued sequences of length $\omega_1$ such that $\{ \alpha : |x_\alpha| > 0 \}$ is countable). The sequence $2 * 1_{\{\alpha\}}$ is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around $1_{[0, \alpha]}$, which is in $c_0(\omega_1)$ because $\alpha < \omega_1$.

I have two (three depending on how you count it) major examples of spaces which have no bad sequences:

- Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.
- Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:
- Any reflexive space: If $U_\alpha$ forms a chain, the sets $F_\alpha = \bigcap_{v \in U_\alpha} \overline{B}(v, r + \epsilon)$ form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in $\overline{B}(c, r + \epsilon)$
- any space with the property that $\textrm{diam}(A) = 2 r(A)$ (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.

So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega_1)$ contain a bad sequence?

I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.

For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.