$\langle\mathbb{R},d\rangle$ is not separable [duplicate]

Let $\langle\mathbb{R},d\rangle$ be a metric space, where for all x,y in $\mathbb{R}$ $$ d(x,y)= \begin{cases}1 & x\neq y\\ 0\quad & x=y \end{cases}$$

what i know about it

• I know that any metric space $\mathbb{R}$ to be separable $\mathbb{R}$ needs to contain a dense countable subset .

• I also know that in case of discrete metric space of $\mathbb{R}$ , a singleton set is infact an open ball of radius 1.

• I also know $\mathbb{R}$ is uncountable set and in this case , $R$ have uncountable open balls.

But how can i conclude by all this then $\mathbb{R}$ is not separable here or $\mathbb{R}$ doesn't contain any countable dense subset .

[ NOTE ] : while the linked question is same but asking area of that user is different . He was asking about why the singleton set is open or closed. And community give response according to it there But it's was understable to me but i want some beyond information's which i already get. Thanks !!

In this metric space every singleton set is open : $\{x\}=B(x,1)$. Hence, every subset is open which makes every subset is closed. So the closure of a countable set is itself and $\mathbb R $ is not countable.

If $D$ is dense in $(\Bbb R,d)$ it must intersect every open ball, i.e. every $\{x\}=B_d(x,1)$. What can you conclude about how many dense sets we have?