Example of sequence of sets $(A_n)$ such that $\liminf{A_n}$ and $\limsup{A_n}$ have a specific property

I need to find an example of a set $X$ and a sequence of sets $(A_n)$ in $\mathcal{P}(X)$ such that $\displaystyle \emptyset \subsetneqq \bigcap_{n} {A_n} \subsetneqq \liminf_{n}{A_n} \subsetneqq \limsup_{n} {A_n} \subsetneqq \bigcup_{n}{A_n} \subsetneqq X$. Can someone give me an idea?

One way to do it is to build what you want by just thinking about what these expressions mean. The first thing you need is for the sequence $\{A_n\}$ of sets to have non-empty intersection, so let's start by stipulating that $1 \in A_n$ for all $n$.
Next, we need there to be some element (other than $1$) that's in all but finitely many $A_n$. So let's stipulate that $2 \in A_n$, $n \geq 2$. Then we have $\liminf A_n = \{1,2 \}$.
Next, we need there to be some element that's in infinitely many of the $A_n$ but not in all but finitely many of them. So let $3 \in A_n$, $n \geq 2$, and $n$ even. Then $\limsup A_n = \{1,2,3 \}$.
Our sequence looks like this so far: $$\{ 1\}, \{ 1,2,3\}, \{1,2 \}, \{ 1,2,3\}, \{ 1,2\},...$$
We're almost done, but now we have $\cup_n A_n = \{ 1,2,3\}$, so let's just add $4 \in A_1$ so that the union is strictly bigger than the $\limsup$. And now we're done: $$\{ 1, 4\}, \{ 1,2,3\}, \{1,2 \}, \{ 1,2,3\}, \{ 1,2\},...$$