$\newcommand{\d}{\mathrm{d}}\newcommand{\M}{\mathcal{M}}\newcommand{\L}{\mathcal{L}^1}$Let $(X,\M,\mu)$ be a finite measure space. The following appears to be an unusual definition - I looked it up and only saw references to Royden's text - so I'll define it here:
Let $\cdot\Delta\cdot$ denote symmetric difference. It can be shown that $A\sim B\iff\mu(A\Delta B)=0$ is a proper equivalence relation, and that $\rho_\mu:\M_{/\sim}\times\M_{/\sim}\to\Bbb R^+$, $\rho_\mu([A],[B])=\mu(A\Delta B)$ is a proper metric. For convenience, denote $\M_{/\sim}$ also by $\M$, and use equivalence classes and distinct sets interchangeably. The metric space $(\M,\rho_\mu)$ is the Nikodym metric space.
I am given:
Let $\{A_n\}\subset\M$ be a sequence converging in metric to the set $A$. Show that $A=\bigcup_{n\ge 1}\bigcap_{m\ge n}A_m$.
This means that $\mu(A_n\Delta A)$ becomes arbitrarily small as $n$ grows large - it is left as an implicit hint from the previous to exercise to observe that $A_n\Delta A=(A_n\cup A)\setminus(A_n\cap A)$.
I couldn't see a way to use this identity however, so I used a different approach modelled of the proof of the completeness of the Nikodym space.
My solution:
$$\rho_\mu(A_i,A_j)=\int_X|\chi_{A_i}-\chi_{A_j}|\,\d\mu\to0$$As $i,j\to\infty$. Then $\chi_{A_n}$ is a Cauchy sequence in $\L(X)$ and by Riesz-Fischer it converges in norm to $f\in\L(X)$ and there is a subsequence $n_k$ on which $\chi_{A_{n_k}}\to f$ pointwise $\mu$-a.e., so $f$ takes the values $0,1$ $\mu$-a.e. If one defines: $$A=\{x\in X:\lim_{k\to\infty}\chi_{A_{n_k}}(x)=f(x)=1\}$$Then $\chi_A=f$ $\mu$-a.e. and thus it follows from the integral that $\rho_\mu(A_n,A)\to0$ as $n\to\infty$. Sidenote - this proves the completeness of the Nikodym metric space.
Now note that $x\in A\iff\exists K\in\Bbb N:\forall k\ge K,\,x\in A_{n_k}$ by construction. Then $x\in\bigcap_{k\ge K}A_{n_k}$ iff. $x\in A$, for some $K$. By taking union over $\Bbb N$, one recovers all possible $K$ such that $x\in A$, so we can write: $$A=\bigcup_{m\ge 1}\bigcap_{k\ge m}A_{n_k}$$
But this is a slightly different expression to the one given as the exercise solution - how can I show that taking a subsequence is unnecessary?
The statement you are asked to prove is in fact not true. For instance, consider Lebesgue measure on $[0,1]$ and let $A=[0,1]$. Let $A_n=[0,1]\setminus I_n$ where $I_n$ is a sequence of intervals whose lengths go to $0$, but such that every point of $[0,1]$ is contained in infinitely many of the $I_n$. (For instance, you could do something like $I_1=[0,1/2],I_2=[1/2,1],I_3=[0,1/3],I_4=[1/3,2/3],I_5=[2/3,1],I_6=[0,1/4],$ etc.) Then $\mu(A\Delta A_n)=\mu(I_n)\to 0$, but $\bigcap_{m\geq n}A_m$ is empty for all $n$ so $\bigcup_n\bigcap_{m\geq n}A_m=\emptyset\neq A$.