Let $(x_{n_{k}})$ be a subsequence of $(x_n)$. Prove rigorously by induction that the indices of the subsequence satisfy $n_{k}\geq k$.
I have the following, and I want to see if this is correct. If not, please make corrections
We will use induction to show that $n_k \geq k$ for all $k$.
(Base Case): Let $k=1$. Since $n_1$ is a postive integer we have $n_1 \geq 1$
(Inductive Step): If $n_k \geq k$ for some positive integer k, then $n_{k+1} \geq k+1$, since by definition of a subsequence $(x_{n_{k}})$, the subsequence indices satisfy $n_k < n_{k+1}$. Thus we have $n_{k+1} \geq k+1 \geq n_k +1$.
Therefore the indices of the subsquence satisfy $n_k \geq k$.
I wasn't going to post an answer, but I do disagree with Riley on some points.
You've done the base case correctly, and it's good to assume, for some $k \in \Bbb{Z}$, we have $n_k \ge k$. That's part of proving any inductive step (if you assumed it for all $k$, that would be a problem!).
Your argument is a bit muddled, but seems to include most of the main ingredients. You are assuming $n_k \ge k$, and that $n_{k+1} > n_k$, as per the definition of "subsequence". From here, I would combine these assumptions to conclude $$n_{k+1} > n_k \ge k \implies n_{k+1} > k.$$ From here, you need to get to the conclusion $n_{k+1} \ge k + 1$, and this is the ingredient that you seem to be missing: $n_{k+1}$ and $k$ are both integers. You need to point out that there is no integer strictly between $k + 1$ and $k$, and therefore an integer like $n_{k+1}$ must be at least $k + 1$.
This might seem a little hand-wavy, because it is, but to show that there is no integer between $k$ and $k + 1$ will take you closer to the axioms than an exercise like this should. My advice would be to assert it as a fact, without proof.