I have the following problem in general topology:
Prove that any open subspace of a separable space is separable.
I know a similar question has been answered here, but mine needs to be proven without metric spaces and open balls, as it's a general topology problem.
Is the following solution sufficient:
Let $O$ be an open subset of the separable space, and let $D$ be the countable dense set in $X$. Consider any open neighborhood $G$ of a point $x \in O$ with the constraint that $G$ be contained in $O$ (since $O$ is open, this can always be done). Since $D$ is dense in $X$; $G \cap O \neq \emptyset $. Therefore, any open neighborhood of a point in $O$ contains an element of $D$. This means $D$ is dense in $O$, therefore $O$ is separable.
Best Answer
It doesn't work as nothing in your proof guarantees that $D \subseteq O$.
Rather, you should show that $D \cap O$ is a dense subset of $O$.
Here is a proof:
You just have to show that $D \cap O$ intersects any non-empty open $U$ of $O$.
So fix such a non-empty open $U$ of $O$. Because $O$ is open, $U$ is also open in $X$ and because $D$ is dense in $X$, we get $U \cap D \neq \emptyset$. But
$$U \cap D = (U \cap O) \cap D = U \cap (D \cap O)$$
so we get $U \cap (D\cap O) \neq \emptyset$, as desired.