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@MonroeEskew No, there's no subtlety involved; it's just that Stone-duality makes it easier (for me) to see what you were asking. The point here is that quotients are dual to subspaces, so you were asking whether the space of uniform ultrafilters has a subspace homeomorphic to $\beta\omega\setminus\omega$. The same reasoning would provide a two-line answer to this question of yours, because the Stone space of that algebra contains a copy of $\beta\omega$, hence of $\beta\omega\setminus\omega$.
What about taking a copy $K$ of $\omega^*$ in the set of uniform ultrafilters on $\omega_1$? The ideal $I=\{A:A^*\cap K=\emptyset\}$ is a uniform and $K$ is the Stone space of $\mathcal{P}(\omega_1)/I$.
Just to be sure: "$A$ separates the points of $A$" should probably be "$A$ separates the points of $S$". And shouldn't $\gamma S$ be $\gamma A$?, the spectrum of the algebra $A$?
@Gaspar You cannot expect an answer depending on the space only: in a countable compact space use an enumeration of the space to find an increasing cover by finite sets. The space is very nice, but this particular cover does nothing useful. Or in the unit interval take $K_n=\{0\}\cup[2^{-n},1]$; you will not capture any compact set that has $0$ as an accumulation point. So, even for compact metrisable spaces there is no guarantee without any assumptions on $\psi$ (which means that some tweaking may be necessary).
I take it that rather than assume discontinuity of $\psi$ we should not assume that it is continuous. The assumption that such a $\psi$ exists does indeed imply $\sigma$-compactness, but the converse also holds: assume $X=\bigcup_{n=1}^\infty K_n$ with each $K_n$ compact and, wlog, $K_n\subseteq K_{n+1}$ always. Define $\psi(x)=\min\{n:x\in K_n\}$; then $\psi$ satisfies your assumption. So the answers that you seek are the answers that one gets for $\sigma$-compact spaces.
