Prove that a subset $A$ of $X$ is dense if and only if every open subset of $ X$ contain some point of $A$ Let $X,τ$ be a topological space. Prove that a subset $A$ of $X$ is dense if and only if every open subset of $ X$ contain some point of $A$
this is what I got
Let $X,τ$ be a topological space
Part 1: Assume that a subset $A$ of $X$ is dense, show that every open subset of $X$ contain some point of $A$
Let $a∈A$, by axiom i) of the closure of set, $A⊂Cl A$, so $a∈Cl A$. Since $A$ is dense, $Cl A=X$, so $a∈X$. Let $ℵ_x $be the the collection of open neihborhood in $X$, by the axiom  ii) of the open neighborhood system,  $a∈X$ then $a∈N$ for each $N∈ℵ_x$. In other word, every open subset of $X$ contain some point of $A$.
Part 2: Assume that every open subset of $X$ contain some point of $A$, show that subset $A$ of $X$ is dense
I'm kinda stuck on how to show $Cl A=X$. I know that I need to show $Cl A⊂X$ and the other way around, but I'm not sure how.

The complement of the closed set $\operatorname{Cl}A$ is open and disjoint to $A$. By assumption the only open set disjoint to $A$ is the empty set.
