Prove that for a subsequence of natural numbers ${n_k}$, $k ≤ n_k$.
What you forgot to mention is that $\{n_k\}_{k=1}^{\infty}$ is a strictly increasing sequence of natural numbers. That's what you need in order to define the subsequence of a sequence.
Fortunately, it is easy to show that this holds for such a sequence. Define the following set:
$$S = \{k \in \mathbb{N}: n_k \geq k\}$$
where $\{n_k\}_{k=1}^{\infty}$ is understood to be a strictly increasing sequence of natural numbers. We know that $1 \in S$. This should be rather obvious. Now, suppose that $k \in S$. Then:
$$n_{k+1} \geq n_k+1 \geq k+1$$
which shows that $k+1 \in S$. So, $S = \mathbb{N}$ and the given result holds.
I mean, you should already be noticing something wrong with the assertion that you have to show that the sequence $k$ converges. That sequence does not converge at all.
Besides, how would that help you show that $\forall k \in \mathbb{N}: n_k \geq k$? The point is that there are a few good ways of reflecting on the ideas that you're coming up with to attack a given problem.
