Separable Subspace of $R^d$ I'd like to show that every $A$ is separable, given $A \subset R^d$ and the usual topology on $R^d$. I know this is true for all metric spaces, but is this true for topological spaces as well?
I've tried a similar approach as in this thread, but couldn't figure out where a property of the metric space is used.
 A: Second countable is hereditary
Any subspace of a second countable topological space is second countable but the same is not true for separable spaces, these concepts coincide for metric spaces or Lidelof spaces.
A space is second countable if and only if there is countable basis $\mathcal{U}$ for its topology.
Suppose that $A \subseteq X$, where $X$ is separable then take $\mathcal{U}$ a countable basis for $X$ and define
$$
 \mathcal{U}_A = \{ U \cap A \mid U \in \mathcal{U} \}
$$
then $\mathcal{U}_A$ is countable, we have to show that it is a basis for its topology. Let $V$ be an open set of $A$, then there is $V'$ an open set of $X$ such that $V = V' \cap A$, as $\mathcal{U}$ is a basis of $X$, $V'$ can be writen as a union of elements $U_i$ in $\mathcal{U}$, and therefore $V = \cup U_i \cap A = \cup (U_i \cap A)$, which is in $\mathcal{U}_A$. Therefore $A$ is second countable.
Separable and second countable metric spaces
Suppose that $X$ is metric then separable and second countable are equivalent. Note that in general you only have that second countable implies separable.
If it is second countable then  the trick is to take an element $x$ for each open of a countable basis and that will form a countable dense subset, conversely, if you have a countable dense subspace a basis consists of open balls of radius $1/n$ around that dense set (here you use the metric). 
Separable is not hereditary
For a counterexample consider the Sorgenfrey line, i.e. $\mathbb{R}$ with the topology generated by sets of the form $[a,b)$ and consider $M$ as the topological product of two copies of the Sorgenfrey line. $\mathbb{Q}\times \mathbb{Q}$ is a dense subset so $M$ is a separable space, but the antidiagonal
$$
\Delta = \{ (x, -x) \mid x \in \mathbb{R}\} \subseteq M
$$
is not separable, in fact it is uncountable and discrete. $(x,-x) \in U = [x,a) \times [-x, b)$, $U$ is open but $U \cap \Delta = \{(x,-x)\}$ meaning that $(x,-x)$ is an isolated point of $\Delta$ and this was done independently of the choice of point in $\Delta$ so it is discrete and the only possible dense subspace is $\Delta$.
