1. The problem statement, all variables and given/known data Let (X,d) be a metric space, A a closed subset of X. Suppose we've found an uncountable subset [itex] U \subseteq A [/itex] such that [itex] \forall x,y \in U, \; d(x,y) \geq c [/itex] for some positive constant c. Show that A is not separable 2. Relevant equations A is separable if it contains a countably dense subset 3. The attempt at a solution For the sake of contradiction assume that A is separable. That is, there exists a countable set S such that cl(S) = A, or [itex] \forall a \in A \; \forall r>0, \; B(a;r) \cap S \neq \emptyset [/itex] Now my idea from here is as follows. Since U is uncountable and S is necessarily countable, we can choose [itex] x \in U\setminus _S [/itex]. Then [itex] \forall r>0 \; B(x;r) \cap S \neq \emptyset [/itex] by assumption. Now I want to show a contradiction using the fact that [itex] \forall x,y \in U , \; d(x,y) \geq c [/itex]. I've been thinking about how to find a suitable r such that [itex] B(x;r) \subseteq U\setminus_S [/itex] which would cause [itex]B(x;r) \cap S = \emptyset [/itex] and give a contradiction. However, I can't find a way to guarantee that a sufficiently small r exists. Hence that idea doesn't seem to be leading anywhere, and I'm not sure what to try next.