Let $\mathcal D$ be an upper semicontinuous decomposition of $\mathbb S^n$ and let $\mathcal D'\subset\mathcal D$ be the set of non-singletons. The decomposition space $^{\mathbb S^n}/_{\mathcal D}$ is homeomorphic to $\mathbb S^n$ if $\mathcal D$ is "shrinkable", and there are many conditions on $\mathcal D$ that ensure this. (For example, there are numerous such conditions in Daverman's book "Decompositions of Manifolds".)
However, the statements I've been able to find assume, in one way or another, that $\mathcal D'$ is countable.
Is there some set of conditions on an upper semicontinuous decomposition $\mathcal D$ of $\mathbb S^n$ that ensures that the decomposition space $^{\mathbb S^n}/_{\mathcal D}$ is homeomorphic to $\mathbb S^n$ (e.g. by ensuring that $\mathcal D$ is shrinkable), but that does not require the set $\mathcal D'$ of non-singletons in $\mathcal D$ to be countable?
(In the situation I have in mind, $\cup_{D\in\mathcal D'}D$ has dense, full-measure complement, each $D\in\mathcal D'$ is homeormorphic to a closed ball in $\mathbb E^k$ for some $k<n$, and $^{\mathbb S^n}/_{\mathcal D}$ is compact, Hausdorff, and second-countable. Moreover, the quotient $\mathbb S^n\to^{\mathbb S^n}/_{\mathcal D}$ has a very nice description. However, my $\mathcal D'$ is definitely uncountable.)