we're asked to prove what is shown in the title (with $$x_n \in X$$), which would prove that the metric space X is not compact. This reminds me of the discrete metric, and my intuition is as follows:
Any subsequence would converge to the same limit. The contrapositive of a theorem states that if a metric space is not Cauchy, then it does not converge.
But for a sequence in a metric space to be Cauchy, $$d(x_n, x_m)$$ should be less than or equal to $$\epsilon$$, but it's 1 whenever $$n \neq m$$, so it seems that $$x_n$$ should not be Cauchy.
Am I going in the right direction here? Any help appreciated, thank you!
Assume that a subsequence $(x_{n_k})_{k \geq 0}$ converges. So it is Cauchy. Then there is $k_0 \geq 0$ such that for $k_1,k_2 \geq k_0$ the inequality holds :$$d(x_{n_{k_1}}, x_{n_{k_2}}) < 1.$$This contradicts the assumption $d(x_{n_{k_1}}, x_{n_{k_2}})=1$.