$A$ is a dense set in $X$ iff $A$ is uncountable

We equip an infinite and uncountable set $$X$$ with the topology $$\mathcal{T}=\left\lbrace U \subset X : U = \emptyset \text{ or } X \setminus U \text{ is countable}\right\rbrace.$$ Prove that $$A \subset X$$ is dense in $$X$$ if and only if $$A$$ is an uncountable set.

I have found a solution for $$(\Leftarrow)$$ by proving its contradiction is wrong. I get an arbitrary open set $$U \neq \emptyset$$, then I have its contradiction $$U \cap A=\emptyset$$, that means $$A \subset X \setminus U (!)$$, because $$A$$ is uncountable but $$X \setminus U$$ is countable.

I have tried to prove $$(\Rightarrow)$$ but I didn't manage to prove it. Is there any hint for it? Thank you!

• Closed sets are exactly $X$ and the countable subsets... – Gae. S. Jul 19 '19 at 14:07

Suppose that $$A$$ is dense, and for the sake of contradiction suppose that $$A$$ is countable. So $$A=X\setminus(X\setminus A)$$ is countable, then $$(X\setminus A)$$ is an open set. Note that $$X\setminus A$$ is not empty, because $$X$$ is uncountable. As $$A$$ is dense, it must happen that $$A\cap(X\setminus A))$$ is not empty, which is a contradiction. So $$A$$ is uncountable.
Hint: By the definition of $$\mathcal{T}$$, the full space $$X$$ is the only uncountable closed set.