I was reading Larsen's proof that if $(X,S,\mu)$ is a positive measure space then $L^\infty(X,S,\mu)$ is isometrically isomorphic to $\mathcal{C}(\Delta(L^\infty(X,S,\mu)))$ and $\Delta(L^\infty(X,S,\mu))$ is zero-dimensional, i.e. has a base of clopen sets in the Gel'fand topology.
The "isometrically isomorphic" part is clear. Then a bijection is shown between measurable sets for $\mu$ and clopen sets in the $\Delta$ space. So $\Delta$ can be endowed with a topology generated by clopen sets whose characteristics are dense in $\mathcal{C}(\Delta)$. The Gel'fand topology makes $\Delta$ compact, hence if I prove the clopen-generated topology is Hausdorff I have they coincide, since clearly the latter is weaker than the former.
Here is how the Hausdorffness is proved.
So suppose $\tau_1,\tau_2\in\Delta$ and $\tau_1\neq\tau_2$. Then there exists some $\hat h\in\widehat{L^\infty}(X,S,\mu)=\mathcal{C}(\Delta)$ such that $\hat h(\tau_1)\neq\hat h(\tau_2)$, as $\mathcal{C}(\Delta)$ separates the points of $\Delta$. WLOG we may assume that $\hat h(\tau_1)\neq0$. It is then easily verified that $\hat f=\hat g\overline{\hat g}$, where:
$$\hat g=\frac{\hat h-\hat h(\tau_2)}{\hat h(\tau_1)-\hat h(\tau_2)}$$
is continuous on $\Delta$ and satisfies $\hat f(\tau_1)=1,\hat f(\tau_2=0,\hat f\geq0$ on all of $\Delta$. Next let $\hat k$ be a finite linear combination of characteristic functions of clopen subsets of $\Delta$ such that $\hat k\geq0$ on all $\Delta$ and $\|\hat f-\hat k\|<\frac13$, which is possible since the characteristics of clopen sets span a norm-dense subalgebra of $\mathcal{C}(\Delta)$ -- which is because the characteristics of measurable sets span a dens subalgebra of $L^\infty$ which is isometric with $\mathcal{C}(\Delta)$. But then it is apparent that $\{\tau|\tau\in\Delta(L^\infty),\hat k(\tau)>\frac23\}$ is a clopen set containing $\tau_1$, whereas $\{\tau|\tau\in\Delta(L^\infty),\hat k(\tau)<\frac13\}$ is a clopen set containing $\tau_2$, and these clopen sets are disjoint. Then $\mathcal{T}$ is Hausdorff.
I changed this quote a little (a little rewording, adding a justification for that norm-denseness, and changing from $L^\infty(X,S,\mu)$ to $\widehat{L^\infty}(X,S,\mu)$ for the image of the Gel'fand transform, and not much more), but the last two sentences are verbatim. $\mathcal{T}$ is the clopen-generated topology.
It is clear to me that those two sets, call the former $A$ and the latter $B$, are open both in the Gel'fand topology and in the clopen-generated topology, since $\hat k$ is continuous w.r.t. the former and a combination of characteristics of clopens in both. It is also clear that they are disjoint, and that $\tau_1\in A,\tau_2\in B$.
But why are they closed?
I mean, I can write them as a union of clopens from the clopen base of the clopen-generated topology, but that does not imply they are closed in the clopen-generated topology, since a union of closed sets is not closed. $A'\{\hat k\geq\frac23\}$ would certainly be closed in both topologies, but $A\smallsetminus A'=\{\hat k=\frac23\}$, and I have no way of knowing whether this is empty or not, or whether a $\tau\in A'$ is or not a limit point of $A$. Or do I?
Given finitely many clopen sets $C_1,\dots,C_n$, consider the $2^n$ sets $D_1,\dots,D_{2^n}$ of the form $$D_j=\bigcap_{l=1}^nE_l,$$where $E_l$ is either $C_l$ or the complement of $C_l$. The sets $D_j$ are disjoint, and every $C_k$ is the union of some collection of $D_j$.
So if $\hat k$ is a linear combination of characteristic functions of clopen sets then it is a linear combination of characteristic functions of disjoint clopen sets. Hence $\hat k^{-1}(A)$ is clopen for any set $A$.