I claim that there are no bad sequences in C(ω<sub>1</sub>). Suppose to the contrary that x<sub>α</sub> is bad. For any countable ordinal β, there is r<sub>β</sub> in C(ω<sub>1</sub>) such that the distance between r<sub>β</sub> and x<sub>α</sub> for α < β is at most 1. For any countable limit ordinal β and any positive rational number ε, there is a smaller ordinal γ < β such that all r<sub>β</sub>(α) are within ε of r<sub>β</sub>(β) for α in [γ,β). For fixed ε, this is a regressive function on the countable limit ordinals. Thus, by Fodor's Lemma, there is a stationary and hence unbounded set of limit ordinals on which the function has constant value, which we may call γ<sub>ε</sub>. Since there are only countably many ε, we may find a countable ordinal γ above all γ<sub>ε</sub>. This ordinal has the property that for all limit ordinals β above γ, we have r<sub>β</sub>(α) = r<sub>β</sub>(β) for all α in the interval [γ,β), since the values are within every ε of each other. That is, every r<sub>β</sub> function is constant from the same fixed γ up to β. Let C<sub>β</sub> be the closed interval of values s such that the constant sequence s of length β lies within 1 of all x<sub>η</sub>(α) for all η ≤ β and all γ ≤ α ≤ β. These are nested and not empty, since r<sub>β</sub>(β) is in C<sub>β</sub>. By compactness, there is a value s in all C<sub>β</sub>. Thus, the number s is within x<sub>η</sub>(α) for all η and all α above γ. Thus, we may form the desired sequence r by using r<sub>γ</sub> up to and including stage γ, augmented with the constant value s at the stages above γ up to ω<sub>1</sub>. This sequence is continuous, and it lies within 1 of every x<sub>η</sub>, as desired. I guess this argument generalizes easily to other C(κ) for ordinals κ having uncountable cofinality, so that Fodor's lemma still holds.