Topology proof: dense sets and no trivial intersection

Question

I was wondering if this proof of this basic topological result concerning the closure works.

Proposition: Let $A \subseteq (X,\tau)$. Then, $A$ is dense in $X$ if and only if every non-empty open subset of $X$ intersects $A$ non trivially.

Proof:
[only if] Assume that $A$ is dense in $X$. This means that $A \cup A' = X$, where $A'$ denotes the set of all the limit points of $A$. Let $G \in \tau$ be non-empty. Thus, $G \subseteq (A \cup A') = X$. Hence, either $G \cap A \neq \varnothing$, which establishes the result, or $G \cap A' \neq \varnothing$, which leads to a contradiction.

[if] Assume that $A$ is not dense in $X$. Thus, there is a $x^* \in X \setminus (A \cup A')$. Notice, that $X \setminus (A \cup A')$ is open, because $A \cup A'$ is closed, and it is not empty. Hence, we can conclude by realizing that $X \setminus (A \cup A') \cap A = \varnothing$, which completes the proof. $\square$

It is one of the first proofs I attempt in topology, and I am not planning to see how it is proved in the book I am reading (unless you answer me that my proof is completely wrong).

Thus, I am really looking forward to any feedback, also contentwise.

Thank you for you time.

Answer 1

I think what you want to say instead is that either $G\cap A\neq 0$ which establishes the result, or else we would have both $G\cap A'\neq 0$ and $G\subseteq X\setminus A$. But since G is open, $G\subseteq X\setminus A\Rightarrow G\subseteq X\setminus \bar{A}$ which contradicts $G\cap A'\neq 0$.

Answer 2

Why does $G\cap A'\ne \emptyset$ leads to a contradiction? We know that $$G\cap A = \emptyset \implies G\subseteq A' - A$$ but you have to say something more in order to reach the absurd (unless you're using some propositions you proved earlier)

The rest is Ok.