Let $(X,d)$ be seperable metric space, then $(Y,d_1)$ where $Y \subset X$ and $d_1$ is induced metric on a subspace $Y$. Then $(Y,d_1)$ is a seperable metric space, too.
I think it is quite easy to prove using the second axiom of choice. My question is that this problem states it is not neccessarily true that when we talk about topology space instead of metric space.
Can anyone explain why and give an counterexample?
Thanks
