I claim that there are no bad sequences in C(ω1).
Suppose to the contrary that xα is bad. For any countable ordinal β, there is rβ in C(ω1) such that the distance between rβ and xα 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 rβ(α) are within ε of rβ(β) for α in [γ,β). (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 γε. Since there are only countably many ε, we may find a countable ordinal γ above all γε. This ordinal has the property that for all limit ordinals β above γ, we have rβ(α) = rβ(β) for all α in the interval [γ,β), (γ,β], since the values are within every ε of each other. That is, every rβ function is constant from the same fixed γ up to β.
Let Cβ be the closed interval of values s such that the constant sequence s of length β lies within 1 of all xη(α) for all η ≤ β and all γ ≤ < α ≤ β. These are nested and not empty, since rβ(β) is in Cβ. By compactness, there is a value s in all Cβ. Thus, the number s is within xη(α) for all η and all α above γ.
Thus, we may form the desired sequence r by using rγfinding a center that works for the sequences up to and including stage γ, using the separability idea in your question, augmented with the constant value s at the stages above γ up to ω1. That is, we solve the problem separately on the first γ many coordinates, and then append the constant s sequence up to ω1. This sequence is continuous, and it lies within 1 of every xη, 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.