Let $X$ be an uncountable set equipped with the topology $\mathcal{T}:= \{\emptyset\}\cup\{A \mid |X \setminus A| \leq |\mathbb{N}|\}$. Let $A \subseteq X$ uncountable. Show that $A$ is dense in $X$.
My attempt:
Let $x \in X$ and suppose that there exists a neighborhood $V$ of $x$ such that $A \cap V = \emptyset$. Let $G \in \mathcal{T}$ with $x \in G \subseteq V$. This means that $X \setminus G$ is at most countable, and hence $X \setminus V$ as well. How to find a contradiction? From this point, I also know that $X = X \setminus A \cup X \setminus V$
If $A \cap V = \varnothing$, then $A \subseteq X \setminus V$.
This cannot be if $A$ is uncountable and $X \setminus V$ is countable.