Another question on almost sure and convergence in probability

Yes, we can take a sequence $\{n_k\}$ which works for almost all $\omega$. To see that, fix a subsequence $\{n_k\}$ such that for each $k$, $$ \Pr\left(|X_{n_k}-X|>2^{-k}\right)\leqslant 2^{-k}.$$ This one can be constructed by induction. Indeed, we first use the definition of convergence in probability with $\varepsilon=1/2$. We know that $\Pr\left(|X_{n}-X|>2^{-1}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that for some $n_1$, $\Pr\left(|X_{n_1}-X|>2^{-1}\right)\leqslant 2^{-1}$. Now assume that we constructed $n_1<n_2<\dots<n_{k-1}$ such that for all $j\in\{1,\dots,k-1\}$, $$ \Pr\left(|X_{n_j}-X|>2^{-j}\right)\leqslant 2^{-j}.$$ We now use the definition of convergence in probability with $\varepsilon=2^{-k}$. We know that $\Pr\left(|X_{n}-X|>2^{-k}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that there is some $N$ such that for all $n\geqslant N$, $\Pr\left(|X_{n}-X|>2^{-1}\right)\leqslant 2^{-k}$. Consequently, a $n_k$ bigger than $n_{k-1}$ and $N$ does the job.

By the Borel-Cantelli lemma applied to $A_k:=\left\{|X_{n_k}-X|>2^{-k}\right\}$, $$P\left(\limsup_{k\to+\infty}\left\{|X_{n_k}-X|>2^{-k}\right\}\right)=0.$$ This proves convergence almost everywhere of $\{X_{n_k}\}$ to $X$.

Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.