# separable subspace

I just want to be sure because I'm a little confused about something.

If I have a topological space $(X,\tau )$ and a subset $A\subseteq X$,

What does it mean that $A$ is separable?

I know that it means that there is a countable subset $S\subset A$ s.t $\overline{S} =A$ but I'm not sure if I need to take the closure with respect to $\tau$ or the closure with respect to the subspace topology.

So you should take the closure within $A$ with the subspace topology.
A bit of thought, however, should convince you that the condition in this case is the same as requiring $\overline S \supseteq A$ where the closure is taken in $X$ -- which may be conceptually easier to imagine.
This is a good question and the answer is that the closure must be taken with respect to $$\tau$$, and not with respect to the subspace topology. Indeed, if $$(X, \tau)$$ is a topological space given any subset $$A\subset X$$, $$A$$ is both open and closed in the subspace topology $$A$$ inherits from $$(X; \tau)$$, hence the closure of $$A$$, in this topology, is $$A$$ itself.