Subspace topology and discrete space Does there exist $A \subseteq \mathbb{R}^{2}$ with usual topology and $A$ not enumerable such that $\tau_{A}$ the subspace topology in $A$ it is a discrete space?

HINT: $\Bbb R^2$ has a countable base $\mathscr{B}$. If $\langle A,\tau_A\rangle$ is a discrete space, then for each $x\in A$ there is a $B_x\in\mathscr{B}$ such that $B_x\cap A=\{x\}$. The map $A\to\mathscr{B}:x\mapsto B_x$ is injective (one-to-one), so ... ?
