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.

Since every (non-empty) open subset of $X$ contains a point from $A$ it has to hold that 
(1.) $\forall U \in \mathcal{O}(X)\setminus\{\emptyset\}: U \cap A \neq \emptyset$. 
Because of $A \subseteq \mathcal{Cl}A$ it holds also that 
(2.) $\complement \mathcal{Cl}A \cap A = \emptyset$
So let's take a look at the given complement:
\begin{align*}
\complement\mathcal{Cl}{A} &= \complement\bigcap_{U\in\mathcal{O}(X),A\subseteq \complement U}\complement U 
\\&= \bigcup_{U\in\mathcal{O}(X),A\subseteq \complement U} U =: U_0.
\end{align*}
By definition, $U_0$ has to be an open set. By (2.) it follows that $U_0\cap A = \emptyset$. Then with (1.), $U_0$ has to be the empty set. 
Therefore $\complement \mathcal{Cl}A = \emptyset$ and so $\mathcal{Cl}A = X$. 
