TITLE: Prove: If a sequence converges, then every subsequence converges to the same limit.
QUESTION [56 upvotes]: I need some help understanding this proof:
Prove: If a sequence converges, then every subsequence converges to the same limit.
Proof:
Let $s_{n_k}$ denote a subsequence of $s_n$. Note that $n_k \geq k$ for all $k$. This easy to prove by induction: in fact, $n_1 \geq 1$ and $n_k \geq k$ implies $n_{k+1} > n_k \geq k$ and hence $n_{k+1} \geq k+1$.
Let $\lim s_n = s$ and let $\epsilon > 0$. There exists $N$ so that $n>N$ implies $|s_n - s| < \epsilon$. Now $k > N \implies n_k > N \implies |s_{n_k} - s| < \epsilon$. 
Therefore: $\lim_{k \to \infty} s_{n_k} = s$.

What is the intuition that each subsequence will converge to the same limit
I do not understand the induction that claims $n_k \geq k$

REPLY [12 votes]: About the indices.
$n_1\geq 1$, because $(n_k)$ is defined to be a strictly increasing sequence of indices.
Assume $n_k\geq k$ for some $k\in\mathbb{N}$, what follows by definition is that $n_{k+1} > n_k\geq k$.
If $m,n\in\mathbb{N}$ such that $m>n$, then $m\geq n+1$.  (property of natural numbers)
We now conclude that $n_{k+1}\geq n_k +1\geq k+1$