Uncountable Discrete Space is not Separable

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space where $\tau$ is the discrete topology on $S$.

Let $S$ be an uncountable set, thereby making $\tau$ the uncountable discrete topology on $S$.

Then $T$ is not separable.

Proof
Let $T = \left({S, \tau}\right)$ be the uncountable discrete topology on $S$.

Let $H \subseteq S$ be everywhere dense in $T$.

Then by definition of everywhere dense, $H^- = S$ where $H^-$ denotes the closure of $H$.

However, as $T$ is a discrete space, $H^- = H$ from Interior Equals Closure of Subset of Discrete Space.

So $H^- = S \implies H = S$.

But $S$ is uncountable.

So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense.

Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable.

Also see

 * Countable Discrete Space is Separable