[Math] Prove that $\langle\mathbb{R}$, discrete metric$\rangle$ is not separable

Question

Let $\langle\mathbb{R},d\rangle$ be a metric space,
where for all x,y in $\mathbb{R}$ $$ d(x,y)= \begin{cases}1 & x\neq y\\ 0\quad & x=y \end{cases}$$
We know that in order for the metric space $\mathbb{R}$ to be separable $\mathbb{R}$ needs to contain a dense countable subset, say $D$.
What we mean by denseness is that if we pick any point $x\in \mathbb{R}\quad$ and an open set $O$ containing it, then $O$ must intersect with $D$.
In a discrete metric space, the singleton set $\{x\}$ is open(Why?). The only way this set can have non-empty intersection with $D$ is if we have $x\in D\quad$. This means the only dense subspace of the discrete space $\mathbb{R}$ is itself $\mathbb{R}$ , which is not countable(Why?).
Thus it is not separable.
Should I provide the reasons to "Whys" in the proof, or is the proof okay? How should I improve it more?

Answer 1

The proof is fine. Singleton sets like $\{x\}$ are open in discrete spaces by definition. The fact that $\Bbb R$ is uncountable is often proved with Cantor's diagonal argument. How much detail you need to add depends on who you are trying to convince.