Theorem: If $X_{n} \to X$ in probability, then for every subsequence $X_{n(m)}$, there exists a further subsequence $X_{n(m_{k})}$ that converges almost surely to $X$.
In Durrett's Probability: Theory and Examples (pg. 65 in the fourth edition) the proof of this says
Let $\varepsilon_{k}$ be a sequence of positive numbers $\downarrow 0$. For each $k$, there is an $n(m_{k})>n(m_{k-1})$ so that $P(|X_{n(m_{k})}-X|> \varepsilon_{k}) \leq 2^{-k}$. Since $$\sum_{k=1}^{\infty} P(|X_{n(m_{k})}-X|>\varepsilon_{k}) < \infty$$ the Borel-Cantelli lemma implies $P(|X_{n(m_{k})}-X|>\varepsilon_{k} > \text{ i.o.})=0$, that is, $X_{n(m_{k})} \to X$ a.s.
My question is: why is it necessary that $\varepsilon_{k}$ be a sequence of positive numbers $\downarrow 0$? I am looking for someone to explain this part of the proof.
The Problem is that the sequence $m_k$ depends on the choice of $\varepsilon$. So for each $\varepsilon > 0$ you would have a different sequence $n(m_k(\varepsilon))$ so that $P(|X_{n(m_k(\varepsilon))} - X| > \varepsilon \text{ i.o.}) = 0$.
If you choose $\varepsilon_k \searrow 0$, then for each $\varepsilon > 0$ you can find an index $K$ so that $\varepsilon_k < \varepsilon$ for all $k \ge K$. Now this implies
$$P(|X_{n(m_k)} - X| > \varepsilon \text{ i.o.}) \le P(|X_{n(m_k)} - X| > \varepsilon_k \text{ i.o.}) = 0,$$
since there are only finitely many $k$ where $\varepsilon_k > \varepsilon$.