Let $X$ be a compact (not necessarily metrizable) topological space. Let $\mathcal{B}$ be its algebra of clopen sets and let us assume that $\mathcal{B}$ is atomless, i.e. every non-empty clopen subset of $X$ can be partitioned into two disjoint non-empty clopen subsets.
This happens e.g. when $X$ is totally disconnected. In that case, $X$ is an inverse limit of a system of finite spaces. However, assume that $X$ is not totally disconnected. Can $X$ be expressed as an inverse limit of compacts with finitely many connected (necessarily clopen) components?
The assumption that $\mathcal{B}$ is atomless is totally irrelevant here. Every compact Hausdorff space $X$ is an inverse limit of compact Hausdorff spaces with finitely many components. To prove this, first note that we may assume $X$ is a subspace of $[0,1]^I$ for some $I$. Say a subset $\prod_{i\in I}U_i\subseteq[0,1]^I$ is a box if each $U_i$ is an open interval and $U_i=[0,1]$ for all but finitely many $i$. Every open set in $[0,1]^I$ a union of boxes, so every closed set (in particular, $X$) is an intersection of complements of boxes. Moreover, it is easy to see that a finite intersection of complements of boxes has finitely many connected components. So now consider the inverse system consisting of all finite intersections of complements of boxes in $[0,1]^I$ that contain $X$ (and their inclusion maps). This is an inverse system of compact Hausdorff spaces with finitely many connected component whose limit is $X$.