Exercise 2.D (Bartle): Let $(A_n)$ be a sequence of subsets of a set $X$. If $A$ consists of all $x \in X$ which belong to infinitely many of the sets $A_n$, show that \begin{equation*} A = \bigcap_{m=1}^\infty\biggr[\bigcup_{n=m}^\infty A_n\biggr] \end{equation*}
Proof: Notice that if $a \not \in A$, then $a$ is only in finitely many $A_n$. Thus, there exists a largest $N \in \mathbb{N}$ such that $a \not \in A_N$. That is, $a \not \in A_n$ for all $n > N$. However, this is true if and only if $a \not \in \bigcap_{m=1}^\infty[\bigcup_{n=m}^\infty A_n]$. Thus, equality holds.
This is valid correct? I have not really seen many double contrapositive proofs, and it trivializes this exercise, so I feel that I might have missed something.
Hint: see if proving this is any easier $$\bigcap_{m=1}^\infty \bigcup_{n=m}^\infty A_n = \{x : \forall m \in \mathbb N, \exists n\geq M \text{ such that } x\in A_n\}$$