\Let $\{a_n\}$ be a sequence such that for some $\epsilon >0$, $$ \lvert a_n-a_m\rvert \ge \epsilon$$ for all $n\neq m$. Prove that $a_n$ has no convergent subsequence.
My thoughts on this is to prove that $a_n$is unbounded and therefore it has no convergent subsequence.
First, I am able to prove $a_n$ is not convergent since $a_n$ is not a Cauchy sequence. Then, $\lvert a_n-L+L-a_m\rvert=\lvert a_n-L\rvert+\lvert a_m-L\rvert \geϵ$, $\lvert a_n-L\rvert\gtϵ/2$; so, $a_n$ is not bounded.
I feel that my proof is not sufficient. Can anyone help me?
If there is any subsequence $(a_{n_{k}})$ of $(a_{n})$ that converges, then for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|a_{n_{k}} - a_{n_{l}}| < \varepsilon$ for all $k,l \geq N$; this contradicts the hypothesis.