Convergence a.s. from convergence in probability (Steele's book)

Question

In Steele's Stochastic Calculus and Financial Applications, on top of p. 114:

The highlighted portion is an argument for surmising almost sure convergence from convergence in probability by passing to a subsequence. I must be missing something, because I don't see how this argument wouldn't imply more generally, meaning that the two modes of convergence are equivalent (which they, of course, are not).

Answer 1

It was proved that $A_n, B_n$ and $C_n$ converges in probability to $I_1, I_2$ and $0$ correspondly.

Let us take $n_k$ such that $A_{n_k}$ converges to $I_1$ a.s. It's possible - see, e.g., math.stackexchange.com/questions/222264
As $B_{n_k}$ converges to $I_2$ in probability we may take $B_{n_{k_j}}$ from $B_{n_k}$ such that $B_{n_{k_j}} \to I_2$ a.s. Finally we will take ${n_{k_{j_l}}} $ such that $C_{n_{k_{j_l}}} \to 0$ a.s.