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? 
 A: 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).
A: Another example is this；

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.
A: 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.
