Let $(M,d)$ be a metric space and $\{x_n\}_{n=1}^\infty\subset M$ be a sequence. Prove that $\forall n\in\mathbb N,\quad\exists \varepsilon> 0 \;B(x_n,\varepsilon)\cap \{x_n\}_{n=1}^\infty = \{x_n\}$
Any hint? I don't know how to start this proof. Maybe reductio ad absurdum? Counterexamples?