If $(a_n)$ is a convergent sequence in a metric space $(X,d)$ and $(a_{k_n})$ is a subsequence, then $(a_{k_n})$ is convergent.
Lemma: $k_n \ge n, \forall n \in \mathbb{N}$
Base case: Let $n=0$. As $k(0) \in \mathbb{N}$, then $k(0) \ge 0$.
Assume that for some natural number $m \ge 0$ that $k_m \ge m$. Then $k_{m+1} > k_m \ge m$, and thus $k_{m+1} \ge k_m +1 \ge m+1 \Rightarrow k_{m+1} \ge m+1$.
Therefore, by induction, $k_n \ge n, \forall n \in \mathbb{N}$.
Proof. Suppose $(a_n) \to l$, where $l \in X$, and let $\varepsilon >0$. Then $\exists N \in \mathbb{N} \ni \forall n \in \mathbb{N}$ we have $|a_n-l|<\varepsilon$ whenever $n \ge N$. By the lemma, $k_n \ge n, \forall n \in \mathbb{N}$, and hence $k_n \ge n \ge N, \forall n \ge N \Rightarrow |a_{k_n}-l|<\varepsilon$.
Therefore, $(a_{k_n}) \to l$.
Corollary: If $(a_n)$ is a sequence in a metric space $(X,d)$ and $(a_{k_n})$ is a divergent subsequence, then $(a_n)$ is divergent.
Proof. Suppose $(a_n)$ is convergent. Then, as shown above, $(a_{k_n})$ is a convergent subsequence.
