Prove $\limsup{A_n}\backslash\liminf{A_n}=\limsup{(A_{n+1}\backslash A_n)}$ $\newcommand{\N}{\mathbb{N}}$
Problem: 
Prove $\limsup{A_n}\backslash\liminf{A_n}=\limsup{(A_{n+1}\backslash A_n)}$
Attempt(Revised): (I am not sure if it's correct. I would appreciate if anyone can help check..)
Let $x\in\left(\limsup A_{n}\right)\backslash\left(\liminf A_{n}\right)$.
First, observe that 
\begin{align}
\left(\limsup A_{n}\right)\backslash\left(\liminf A_{n}\right) & =\left(\limsup A_{n}\right)\cap\left(\liminf A_{n}\right)^{c}\\
 & =\left(\limsup A_{n}\right)\cap\left(\limsup A_{n}^{c}\right)
\end{align}
So, we have for any $n\in\N$ 
\begin{align}
x\in\limsup A_{n} & \Rightarrow\exists m_{1}>n\mbox{ s.t }x\in A_{m_{1}}\\
x\in\limsup A_{n}^{c} & \Rightarrow\exists m_{2}>n\mbox{ s.t }x\in A_{m_{2}}^{c}
\end{align}
We can assume that $m_{2}\geq m_{1},$ since if not we can acquire
another $m_{2}>m_{1}$ using the definiton. Suppose that $x\in\cap_{k>m_{2}}A_{k}^{c}$,
then $x\notin\limsup A_{n}$, which contradicticts with our choice
of $x$. Therefore $x\in A_{k}$ for some $k>m_{2}.$ By well-ordering
principle, we can acquire the smallest $k$ which satisfy the requirement.
Thus, we have $x\in A_{k}\backslash A_{k-1}=A_{\left(k-1\right)+1}\backslash A_{k-1}$.
So upon reindexing, we have for $\forall n\in\N,$ $\exists k\geq n$
such that $x\in A_{k+1}\backslash A_{k}$. Therefore, $x\in\limsup A_{k+1}\backslash A_{k},$which
implies $\limsup A_{n}\backslash\liminf A_{n}\subseteq\limsup A_{k+1}\backslash A_{k}$.
On the other hand, let $x\in\limsup\left(A_{n+1}\backslash A_{n}\right)$.
Then for any $n\in\N,$ $\exists m\geq n$ such that $x\in A_{m+1}\backslash A_{m}$,
i.e., $x\in A_{m+1}$ and $x\in A_{m}^{c}.$ Since $n$ is arbitrary,
we have 
\begin{align}
x & \in\left(\limsup A_{n}\right)\cap\left(\limsup A_{n}^{c}\right)\\
 & =\left(\limsup A_{n}\right)\cap\left(\liminf A_{n}\right)^{c}\\
 & =\left(\limsup A_{n}\right)\backslash\left(\liminf A_{n}\right)
\end{align}
hence, $\limsup\left(A_{n+1}\backslash A_{n}\right)\subseteq\left(\limsup A_{n}\right)\backslash\left(\liminf A_{n}\right)$.
Therefore, we have 
\begin{align}
\limsup\left(A_{n+1}\backslash A_{n}\right) & =\left(\limsup A_{n}\right)\backslash\left(\liminf A_{n}\right)
\end{align}
as desired. 
 A: Taking $m_1$ to be the smallest $m>n$ such that $x\in A_m$ does not guarantee that $x\notin A_{m_1-1}$. Suppose that $x\in A_n\cap A_{n+1}$; then $m_1=n+1$, and $x\notin A_{m_1}\setminus A_{m_1-1}$. However, you can argue as follows. 

For each $n\in\Bbb N$ we know that $x$ belongs to exactly one of $A_n$ and $X\setminus A_n$ (where the sets $A_k$ are subsets of $X$). Suppose that there us some $n\in\Bbb N$ such that for each $k\ge n$, if $x\notin A_k$, then $x\notin A_{k+1}$. Then either $x\in A_k$ for all $k\ge n$, or there is an $m\ge n$ such that $x\in X\setminus A_m$, in which case an easy induction shows that $x\in X\setminus A_k$ for all $k\ge m$. In the first case $x\notin\limsup_k(X\setminus A_k)$, and in the second case $x\notin\limsup_kA_k$, contradicting the choice of $x$. Thus, for each $n\in\Bbb N$ there is a $k\ge n$ such that $x\in A_{k+1}\setminus A_k$, and it follows that $x\in\limsup_k(A_{k+1}\setminus A_k)$.

The other direction is fine.
You don’t need to separate the case $\limsup_kA_k\setminus\liminf_kA_k=\varnothing$: once the first one is fixed, the previous arguments show that that
$$\limsup_kA_k\setminus\liminf_kA_k\subseteq\limsup_k(A_{k+1}\setminus A_k)$$
and
$$\limsup_kA_k\setminus\liminf_kA_k\supseteq\limsup_k(A_{k+1}\setminus A_k)$$
irrespective of whether the lefthand side is empty. When you suppose that $$x\in\limsup_kA_k\setminus\liminf_kA_k\;,$$ you’re not actually requiring $\limsup_kA_k\setminus\liminf_kA_k$ to be non-empty: you’re merely about to show that if it has some element $x$, then that element also belongs to $\limsup_k(A_{k+1}\setminus A_k)$.
