I need to proof that if a deleted epsilon neighborhood contains at least one element, it must contain an infinite number of elements. So, this is my proof. Am I doing things wrong? I will try to reach a contradiction.
Consider an deleted epsilon neighborhood of $s_0$: $N_{\epsilon}^-(s_0)$ and define the set S to be the set containing all points of this neighborhood. Suppose, S contains not an infinite number of elements. So S would contain a finite number of elements. Define a new set: $D=\{|s-s_0| | s \in S\}$. Since S was finite, this set is finite and therefore we could pick the smallest element of it. So, there exists an $s^* \in S$ such that $|s^*-s_0|$ is the smallest element of D. Since we know the neighborhood contains at least one element, call it $s_1$, we now that for any $\epsilon$ holds that $|s_0-s_1|<\epsilon.$ Now, pick $\epsilon = \frac{1}{2}|s^*-s_0|$, then $|s_0-s_1|\ge|s^*-s_0|>\frac{1}{2}|s^*-s_0|=\epsilon$ which contradicts the fact that $|s_0-s_1|<\epsilon$. So the deleted epsilon neighborhood of $s_0$ must contain an infinite number of elements.
Regards, Kevin