Prove the existence of an uncountable set of points whose pairwise distances are $>\epsilon$ means a metric space is not separable

Question

Prove that a metric space is not separable if $\exists \epsilon > 0$ such that there exists an uncountable set such that the pairwise distances between the points of this set are greater than $\epsilon$

My attempt: Let $Y$ be a metric spacae. Suppose $Y$ is separable. Then $\exists E$, a countable, dense subset of $Y$. $E$ is dense implies $\forall e \in E \, \exists y \in Y$ s.t. $\forall \epsilon>0, |y-e|<\epsilon$. I'm trying to reach a contradiction about how $E$ can't be countable, but I can't quite get there

Answer 1

Take the uncountable set, and for each element in that set take the open ball centered at that point with radius $\varepsilon/2$. These balls are all pairwise disjoint, and if the space is separable then any countable dense subset must have at least one point in each ball.

Answer 2

Let $E = (e_n)_{n \in \mathbb{N}}$ be the countable dense set and let $A$ be the uncountable set in question. Define a map $f : A \to \mathbb{N}$ as follows

$$f(y) = \min\left\{n \in \mathbb{N} : d(y, e_n) < \frac\varepsilon2\right\}$$

for $y \in A$.

Notice that $f$ is injective:

$f(x) = f(y) = n$ implies that $d(e_n, x) < \frac\varepsilon2 $ and $d(e_n, y) < \frac\varepsilon2 $ so

$$d(x,y) \le d(e_n, x) + d(y, e_n) < \varepsilon$$

which implies $x = y$ since $x, y \in A$.

Therefore $\operatorname{card} A \le \operatorname{card}\mathbb{N}$ which is a contradiction with the assumption that $A$ is uncountable.