# Is uncountable subset of separable space separable?

I have to prove that any uncountable $B\subseteq \mathbb{R}$, where $(\mathbb{R},\epsilon^1)$ is euclidean topology and topology on B is relative, is separable. And I know it's true because every subset of separable metric space is separable.

But what if we are given separable space $(X,\tau)$, $X$ uncountable, and $A \subseteq X$ uncountable subset with relative topology. Is $(A,\tau_A)$ separable and if it is, how to prove it?

Consider the Niemytzki (or Moore) plane. This space is separable (the family of points with both coordinates rational is dense), but the $x$-axis $A = \{ \langle x , 0 \rangle : x \in \mathbb{R} \}$ is an uncountable closed discrete subset (and so $A$ with the subspace topology is discrete, and is therefore not separable).
Let $\omega$ denote infinite countable cardinal and $\mathcal E \subset [\omega]^\omega$ is a maximal almost disjoint family, where $[\omega]^\omega = \{ A \subset \omega: |A| = \omega \}$. Let $\Psi(\mathcal E)$ denote the topological space whose point-set is $\omega \cup \mathcal E$, with the topology generated by isolating each $\alpha \in \omega$, and the basic nbhds about $E \in \mathcal E$ are all sets of the form $\{E\}\cup (E\setminus F)$, where $F \in [E]^{< \omega}$. This we can called $\Psi$ space.
Claim: It is separable. However the subspace $\mathcal E$ of $\Psi$ is uncountable closed discrete, and hence it is not separeblae.
The Sorgenfrey Plane is another example. It's a separable space, but the antidiagonal line$\{⟨x,-x⟩:x\in \mathbb{R}\}$ is discrete and uncountable.