Let $(\mathcal{M}, \mathcal{O}, \mathcal{A})$ be a smooth manifold with boundary, where $\mathcal{M}$ is a set, $\mathcal{O}$ is a topology and $\mathcal{A}$ is a smooth atlas. Under which conditions its boundary is the countable union of smooth manifolds?.
There exist a simple example where the boundary of a smooth manifold is not the countable union of smooth manifolds?
This depends on the definition of manifold you are working with. If you look this up in Wikipedia you will see the requirement that a manifold, by definition, is second countable (has a countable base for it's topology). If you take over this requirement to the definition of a manifold with boundary, then this implies that the number of boundary components of a manifold with boundary is at most countable.
The requirement of second countability may be omitted. In appendix A of Spivaks "Comprehensive Indroduction to Differential Geometry", volume 1 (which is worth reading if you are interested in what may happen if you drop paracompactness from the definition), it is shown that the following requirements are equivalent for a manifold $M$ (which, in this appendix, is defined as a Hausdorff space which is locally homeomorphic to some Euclidean space):
So, quite obviously, if the number of components is uncountable (there is no reason why this should not be true with this definition), and if every component is a manifold with boundary, then the answer to your question is no.
Note: I'm a bit sloppy here with the destinction between manifold and manifold with boundary. The definition of manifold typically implies that the boundary -- in the sense of the definition of a manifold with boundary -- is empty. It should be clear, however, that the definition of a manifold with boundary carries over to both the cases where you assume that the manifold is second countable or not.