My first question on this point was not answered. Here is the first part of Shilov's proof of Baire's theorem (not an exact lift from the book as I avoided mathematical symbols). I am trying to be more explicit here so perhaps I will get an answer.
THEOREM: Suppose a complete metric space is the union of a countable number of closed subsets $F_1, F_2 \ldots \subset M$. Then at least one subset $F_n$ contains a closed ball in $M$.
PROOF: Suppose to the contrary that none of the sets $F_1, F_2, \ldots$ contains a closed ball, and let $X_1$ be a point not belonging to $F_1$.
(First question: Why does $M$ need to have any points whatsoever? In which case there are no points not belonging to $F_1$.)
Since $F_1$ is closed there is a closed ball surrounding $X_1$ that does not intersect $F_1$. And within one half the radius of this closed ball there is a point $X_2$ not belonging to $F_2$. ETC ETC ETC
(Second Question: Why can't a complete metric space have only one single point $X_1$? Why does it need to have this second point?)
Answer to both your questions: $M$ doesn't need to be non-empty or contain more than two points. The thing is that the Baire category theorem is uninteresting if $M$ doesn't contain infinitely many points, since for such spaces the only dense set is the whole space itself. So, in the proof one only needs to treat the case where $M$ is an infinite set.
Edit: This is not to imply of course, one can just leave out "uninteresting" cases. What I was trying to say is that in these cases only the whole space $X$ is dense. So if we have a collection of dense sets and we take the intersection, we end up with $X$ again, which is of course dense in itself. So the theorem holds trivially.