Essentially you are to prove that the two sets are equal. To prove that two sets are the same is to prove that every element in one set is in the other and vice versa.
If $x\in A$, then for every $n$: $x\in\cup_{m=n}^{\infty}A_m$, i.e. for every $n$ there exists some $m\ge n$ such that $x\in A_m$. Now, for $n=1$, this means that there is $m\ge 1$ such that $x\in A_m$, hence, $x$ is contained in at least one $A_m$. If there were only finite number of $A_m$'s containing $x$, then there were the maximum index $m=M$ such that $x\in A_m$. But then for $n$ equal to $M+1$ there is another $m\ge M+1>M$ such that $x\in A_m$, and $M$ is not the maximum index. Contradiction. Hence, there are infinitely many sets containing $x$.
If $x$ is contained in infinitely many sets, then for every $n$ there is some $m\ge n$ such that $x\in A_m$. Hence, for every $n$: $x\in\cup_{m=n}^{\infty}A_{m}$, and $x\in A$.