Proving that if $(X,d)$ is a complete and infinitely countable metric space, then there is $x \in X$ such that $\{x\}$ is open [closed]

I tried proving it as follows:

Suppose for a contradiction that all $x \in X$ are not open, since the metric space is countable, this means $X = \cup_n {x_n}$. Because every singleton is not open and closed, the interior of the closure is empty. This means each singleton is a nowhere dense set, which contradicts Baire Category Theorem (that a complete metric space is not Meager).

Is this a valid proof?