First of all, you should use the fact that the sequence $a_k^*=\sup\{a_n; n\ge k\}$ converges to $+\infty$. (Since $\limsup a_n =\lim_{k\to\infty} \sup\{a_n; n\ge k\}$.) The same is not true about $\inf\{a_n; n\ge k\}$.
What we know about the sequence $a_k^*$.
- It is non-increasing: $a_k^*\ge a_{k+1}^*$.
- Since it is non-increasing, we know that $\lim_{k\to\infty} a_k^* = \inf_k a_k^*$. * The equation $\inf_k a_k^*=\infty$ implies that each $a_k^*$ is equal to $+\infty$.
If supremum of some set of real numbers is $+\infty$, then it contains arbitrarily large numbers. So we can construct $a_{n_k}$ inductively as follows: If we know $a_{n_1},\dots,a_{n_k}$, then we choose $a_{n_{k+1}}$ in a such way that:
- $n_{k+1} > n_k$;
- $a_{n_{k+1}} \ge k+1$.
We know existence of such $n_{k+1}$ from the fact that $\sup\{ a_n; n>n_k\}=+\infty$.
Maybe it is worth mentioning that this result is true not only for $+\infty$. If $S=\limsup a_n$, then there is a subsequence $(a_{n_k})$ such that $S=\lim_{k\to\infty} a_{n_k}$.