Intuitively, my question is: how many times do we have to mod out by an closed equivalence relation with all classes compact in order to collapse Baire space $\omega^\omega$ to a singleton?

In more detail, given a space $\mathcal{X}$ let $\mathsf{CoLe}(\mathcal{X})$ be the least $\theta$ such that there is a sequence of spaces $(\mathcal{X}_\eta)_{\eta<\theta+1}$ where

• $\mathcal{X}_0=\mathcal{X}$,

• each $\mathcal{X}_{\eta+1}$ is the quotient of $\mathcal{X}_\eta$ by a closed equivalence relation each of whose classes is compact in the sense of $\mathcal{X}_\eta$,

• for $\lambda<\theta$ a limit ordinal, the space $\mathcal{X}_\lambda$ is the colimit of the family of $\mathcal{X}_\eta$s with $\eta<\lambda$, and

• $\mathcal{X}_\theta$ is a singleton.

For example, if we use $\mathbb{R}$ instead of $\omega^\omega$ then the corresponding ordinal is $\omega$: at stage $n$ we can collapse $[-n,n]$ to a point, and at stage $\omega$ this gives us the one-element space. Similarly, by collapsing just a pair of points at a time, we clearly have an upper bound of $\mathfrak{c}$, and it's not hard to show that $\omega_1$ is a lower bound for $\mathsf{CoLe}(\omega^\omega)$ (essentially this is an elaboration on the non-$\sigma$-compactness of $\omega^\omega$).

However, beyond that things aren't clear to me. In particular:

Is it consistent with $\mathsf{ZFC}$ that $\mathsf{CoLe}(\omega^\omega)<\mathfrak{c}$?