Prove that a complete metric space contains a non-empty open set

Question

In the proof of Baire's Category Theorem in the book by Kreyszig, it is mentioned that a complete metric space will contain a non-empty open set. I like to know the proof of this.

Answer 1

There does not exist a metric space $(X,d)$ where $|X|>1$, in which every open subset is empty.

Every metric space is a Hausdorf space thus if there existed such space then for any two distinct points $x,y \in X$ we would not be able to find two open disjoint sets, which contain $x$ and $y$ respectively,to seperate them.

Thus the existence of such a metric space contradicts the Hausdorf property of metric spaces.

We don't even need the completeness of $X$

Now if $X=\{x\}$ then the only open proper subset of this space is the empty set.

This space satisfies the Baire's theorem because the only dense and open subset of $X$ is the space $X$ itself.

Thus the form of every sequence of open and dense subsets in $X$, is a constant sequence $U_n=\{x\}$ where the interestion is $\{x\}=X$

Note that in this space the empty set is not dense because it is closed thus its closure(as the smallest closed set containing it) is the empty set.

In every metric space the whole space and the empty set are open and closed.