The answer is given by the following theorem (Davidson's *$C^\*$-Algebras by Example*, III, 2.5).

Let $A$ be a commutative C*-Algebra. Then the following statements are equivalent:

- $A$ is an AF-algebra, i.a. a colimit of a sequence of finite-dimensional C*-algebras
- $A$ is separable and the projections in $A$ generate a dense subspace.
- The spectrum of $A$ is totally disconnected.

Thus, the category of locally compact totally disconnected Hausdorff spaces is equivalent to the category of commutative AF-algebras; the inverse functors are $C_0(-,\mathbb{C})$ and $\mathrm{Spm}$. Under this equivalence, compact corresponds to unital. The finite-dimensional commutative algebras are just powers of $\mathbb{C}$, corresponding to finite sets, but colimits produce interesting examples:

For example, if $X$ is the Cantor set, then it is easy to see that $C_0(X,\mathbb{C})$ is the colimit of the sequence $\mathbb{C} \to \mathbb{C}^2 \to \mathbb{C}^{2^2} \to \dotsc$, where $\mathbb{C}^{2^n} \to \mathbb{C}^{2^{n+1}}, a \mapsto (a,a)$. In general, AF-algebras are classified via so-called Bratteli-diagrams (see loc. cit).

By the way, there is a nice connection to Stone duality, which says that $C_0(-,\mathbb{F}_2)$ exhibits an equivalence of categories between the category of locally compact totally disconnected Hausdorff spaces and the category of boolean rings. In the diagram you just exchange $\mathbb{C}$ with $\mathbb{F}_2$. I would love to see a purely algebraic functor $\mathbb{F}_2 \otimes_{\mathbb{C}} (-)$, which doesn't use the spectrum as an intermediate step ...