Counterexample in topology general Let $X$ be complete metric space and $D$ a dense subset countable.
Question: If $D$ is an open and discrete, then $X$ is countable. Even more, if  If $D$ is an open and discrete, then $X\setminus D$ is not perfect.
Any suggestions. Thanks
 A: Unless some mistake eludes me below, it should not hold. Consider the subset $D\subset\mathbb{C}$ by
$$D:=\lbrace (1-2^{-n})e^{i\pi k2^{-n}}|n\in\mathbb{N}^*,0\leq k < 4^n\rbrace$$
and $X\subset \mathbb{C}$, being the union of $D$ with the unit circle
$$X:=D\cup \mathbb{T} $$
You can imagine $D$ as a discrete set of complex numbers, that approaches the unit circle, as shown below:

Give them both the relative topology induced by $\mathbb{C}$. Then, $X$ is a closed subset so it is a complete metric space, and $D$ is a discrete subset, open and dense in $X$. All the conditions above hold but $X$ is uncountable, as it contains the unit circle. Even more, $X\setminus D=\mathbb{T}$ which is perfect.
A: Let  $D=\big\{\big(\frac{1}{n}, \sin(n)\big)\big\}_{n\in\mathbb{N}}\subseteq\mathbb{R}^2$ and define $X=\overline{D}$ to be its closure in $\mathbb{R}^2$.
So $X$ is complete as a closed subset of $\mathbb{R}^2$. $D$ is countable and dense in $X$ by definition. Now $D$ is open in $X$ being a preimage of $(0,\infty)$ via projection map $(x,y)\mapsto x$. It is also discrete because each point is on a line $x=\frac{1}{n}$ (thus we can "widen" those lines into open disjoint neighbourhoods).
What is left to prove is that $X$ is uncountable. For that you have to realize that $\{\sin(n)\}_{n\in\mathbb{N}}$ is dense in [-1,1]. Thus $\{0\}\times[-1,1]\subseteq X$ which completes the counterexample. $\Box$
