Dense subsets

Proposition A subset $Y\subseteq X$ is dense iff for every non-empty open set $O\subseteq$ we have that $Y\bigcap O\ne \varnothing$.

Proof $(⟹)$ Since $Y$ is dense, this menas that the closure of $Y$ is the entire set $X$:

$$\overline{Y}=X$$

pick any non-empty open set $O$, then this is an open neighborhood of one of its elements. By definition of the closure, $O\bigcap Y\ne 0$. $(⟸)$ Suppose that $x\notin \overline{Y}$. Then there exists a neighborhood of $x$ such that $N(x)\bigcap Y=\varnothing$. $\square$