Is a Metric space $(X,d)$ with $X=\{x\}$ an open set?

Question

I've recently started to study functional analysis using "Introduction to functional Analysis" of Edwin Kreyszig. In this book there is a theorem that states that every metric space $X$ is an open set. But what about a metric space that contains only one element? Such space couldn't be a open set because $d(x,x)=0$, and the definition of open set requires $d(x,y)=r$ where $r>0$.

Answer 1

Well, a set $X$ is a metric space when there exists a map $d \colon X \times X \to [0,+\infty)$ such that

1. $d(x,y)=d(y,x)$
2. $d(x,y)=0$ implies $x=y$
3. $d(x,y)\leq d(x,z)+d(z,y)$

for all $x$, $y$, $z \in X$. I can't see any reason why $\operatorname{card}X>1$.