Let $F$ be the field of clopen subsets of the Cantor space $2^\omega$. Then $F$ is countable, atomless, and incomplete.
Question: What does incomplete mean precisely? Does that mean that not every subset of $F$ has a least upper bound (equivalently, a greatest lower bound), or does that mean that none of the subsets of $F$ have a least upper bound (or greatest lower bound)? It is the fact that $F$ is both countable and atomless which confuses me.
Suppose that $O$ is an open set of $2^\omega$ such that $\overline{O}$ is not clopen. (There are plenty of those as we have $\mathfrak{c}$ many open sets but only $\aleph_0$ many clopen ones.).
As the clopens form a base for $2^\omega$ we can write $O = \bigcup \mathcal{C}$ for some countable family of clopen subsets $\mathcal{C}$. Then $\mathcal{C}$ does not have a supremum in $\operatorname{Clop}(2^\omega)$:
Suppose we have a clopen set $D$ that contains all members of $\mathcal{C}$(so an upper bound), then $O =\bigcup{C} \subset D$, and also $\overline{O} \subset D$, by how we chose $O$, $D \setminus \overline{O}$ is non-empty open and so contains some non-empty clopen set $D'$ and then $D \setminus D' $ is a strictly smaller clopen upperbound for $\mathcal{C}$, so no smallest upperbound exists.
This is what the incompleteness means, there are some (necessarily infinite) subsets of $\operatorname{Clop}(2^\omega)$ that do not have a supremum in it.
In $\operatorname{RO}(2^\omega)$ (the completion of $\operatorname{Clop}(2^\omega)$) the supremum of $\mathcal{C}$ is $\operatorname{int}\overline{O}$, of course.