Proof by contradiction
Suppose that $\{X_n\}$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ such that $$\forall N\in\mathbb N,\exists n=n(N) : n>N~~~and ~~~ |X_n -\ell|>\varepsilon_0 $$
For $N_1=1$ there exists $n_1$ such that
$$n_1>N_1 ~~~and ~~~ |X_{n_1} -\ell|>\varepsilon_0 $$
Taking successively $N_{k+1}> \max\{N_k, n_k,k+1\}$ there exists $n_{k+1}>N_{k+1}$ such that,
$$ |X_{ n_{k+1}} -\ell|>\varepsilon_0 $$
It is easy to see that, $\{X_{ n_k}\}_k$ is a subsequence of $\{X_{ n}\}_n$
since
$$ n_k< n_{k+1} \quad i.e ~~\text{the map }~~k\mapsto n_k~~~\text{Is one-to-one}$$
However, $$\forall k,~~ |X_{ n_{k}} -\ell|>\varepsilon_0 \qquad \text{and}~~~\{X_{ n_{k}} \}~~~\text{is bounded} $$
Therefore By Bolzano-Weierstrass Theorem's there exists $\{X_{ n_{k_p} }\}_p$ subsequence of $\{X_{ n_{k} }\}_k$ which converges to some limit $\ell_1 $
but $\{X_{ n_{k_p} }\}_p\to \ell_1$ is also a converging subsequence of $\{X_n\}_n$
By assumption, $\ell=\ell_1$ that is together with the fact $\{X_{ n_{k_p} }\}_p$ is a subsequence of $\{X_{ n_{k} }\}_k$ we have
$$0=\lim_{p\to\infty } |X_{ n_{k_p} }-\ell|>\varepsilon_0>0~~~\text{which is a CONTRADICTION}$$
Note that
$$\forall p,~~|X_{ n_{k_p} }-\ell|>\varepsilon_0$$
Since
$$\forall k,~~|X_{ n_{k}} -\ell|>\varepsilon_0$$
Suppose $a_n$ does not converge to $l$. By definition there exists $\epsilon>0$ such that for every $N$ there exists $n>N$ with $|a_n-l|>\epsilon$. Hence there is a subsequence $a_{n_j}$ with $$|a_{n_j}-l|>\epsilon$$for all $j$.
If the sequence $a_{n_j}$ were convergent we'd be done. There's no reason to think that it's convergent. But it's bounded, so...
Best Answer
$k$ is any positive integer. Then, $(n_k)_{k \geq 1}$ is an indexing of positive integers, which is strictly increasing. So for instance, when you take $n_1$, you can take any available positive integer, so trivially $n_1 \geq 1$. If $n_m \geq m$ for some positive integer m, then since $n_{m+1} > n_m$, we must have $n_{m+1} \geq m+1$ as well. So by induction, $n_k \geq k$ for all $k$.