Can I choose $\varepsilon$ too small for there to be another element in $B(x,\varepsilon)$ in a normed space?

Question

Consider a normed space $X=(X, ||\cdot||_X)$, $x\in X$ and the open ball $B_\varepsilon(x)=\{y\in X:||x-y||_X<\varepsilon\}$ with $\varepsilon>0$ arbitrarily chosen.
Can I be certain that there will be a different element $x\neq y\in X $ s.t. $y\in B_\varepsilon(x)$? And how do I know that?

Potential follow-up question: does this change in a compact subset $M\subseteq X$? And if so, why?

Answer 1

No, for example if you consider $X=\{0,+, \cdot \}$ then $B_{1}(\varepsilon)$ has just one element, the zero. Moreover this normed space is compact.