Try reading it piece by piece. Recall that $A\cup B$ means that at least one of $A$, $B$ happens and $A\cap B$ means that both $A$ and $B$ happen. Infinite unions and intersections are interpreted similarly. In your case, $\bigcup_{k=n}^{\infty}A_k$ means that at least one of the events $A_k$ for $k\geq n$ happens. In other words "there exists $k\geq n$ such that $A_k$ happens".
Now, let $B_n=\bigcup_{k=n}^{\infty}A_k$ to simplify notation a bit. This gives us $\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k = \bigcap_{n=0}^{\infty}B_n$. This is interpreted as "all of the events $B_n$ for $n\geq 0$ happen" which is the same as "for each $n\geq 0$ the event $B_n$ happens". Combined with the above interpretation, this tells us that that $\limsup A_n$ means "for each $n\geq 0$ it happens that there is a $k\geq n$ such that $A_k$ happens". This is precisely the same as saying that infinitely many of the events $A_k$ happen.
The other one is interpreted similarly: $\bigcap_{k=n}^{\infty}A_k$ means that for all $k\geq n$ the event $A_k$ happens. So, $\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$ says that for at least one $n\geq0$ the event $\bigcap_{k=n}^{\infty}A_k$ will happen, i.e.: there is a $n\geq 0$ such that for all $k\geq n$ the event $A_k$ happens. In other words: $\liminf A_n$ is the event that from some point on, every event happens.
Edit: As requested by Diego, I'm adding a further explanation. Sets are naturally ordered by inclusion $\subseteq$. This is a partial order, even a lattice. (Putting aside the fact that the universe of sets is not a set.) In fact, every family of sets has an $\inf$ and $\sup$ with respect to $\subseteq$, which can be defined by: $$\inf_{\lambda\in\Lambda}A_\lambda =\bigcap_{\lambda\in\Lambda}A_\lambda$$ and $$\sup_{\lambda\in\Lambda}A_\lambda =\bigcup_{\lambda\in\Lambda}A_\lambda.$$
Now, the usual definition of $\limsup$ and $\liminf$ (of sequences of real numbers) can be rephrased in terms of infima and suprema as follows: $$\liminf_{n\to\infty}a_n=\sup_{n\geq 0}\inf_{k\geq n} a_n$$ and $$\limsup_{n\to\infty}a_n=\inf_{n\geq 0}\sup_{k\geq n} a_n.$$
We can now use the same definition for sets: $$\liminf_{n\to\infty}A_n=\sup_{n\geq 0}\inf_{k\geq n} A_n$$ and $$\limsup_{n\to\infty}A_n=\inf_{n\geq 0}\sup_{k\geq n} A_n.$$ Rewriting this in terms of $\bigcup$ and $\bigcap$, we get precisely the definitions from the question.
$A_n = \{ \frac{m}{n} ,m \in \mathbb{N}\}$
$\limsup A_n = \{ x | x \in A_n $ for infinitely many $n\}$.
$\liminf A_n = \{ x | x \notin A_n $ only for finitely many $n\}$.
It is clear that no irrational numbers belong to $A_n$ for any $n$. So it is enough for us to consider rational numbers.
$x \in A_n \iff xn=m$ for some $m \in \mathbb{N}$.
If we write $x=\frac{p}{q}$ in lowest form (It's possible that $q=1$), then $xn=m$ for some $m$ if and only if $n | q$. That is, $x \in A_{kq} \forall k \in \mathbb{N}$, since $\frac{p(kq)}{q} = pk \in \mathbb{N}$. Thus, every rational number $x$, $x \in \limsup A_n$.This shows that $\limsup A_n = \mathbb{Q}$.
If $x \in \liminf A_n$, then there is some natural number $N$ such that $x \in A_n$ if $n > N$. Now, we have that $\frac{p(N+1)}{q}$ is an integer, and $\frac{p(N+2)}{q}$ is an integer. Subtracting, we have that $\frac{p}{q}$ is an integer,which happens only when $x$ is an integer itself. On the other hand, of course an integer is in $A_n$ for every $n$ . Hence, it follows that $\liminf A_n = \mathbb{N}$.
I'm keeping my explanation on, since you are attempting to understand it.
$\limsup A_n =\displaystyle\bigcap_{n=1}^\infty \bigcup_{k \geq n} A_k$.
$\liminf A_n =\displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$.
$\forall i \in \mathbb{R}-\mathbb{Q }$, $i \notin A_n \ \forall n$, so $i \notin \limsup A_n,i \notin \liminf A_n$.
Let $x=\frac{p}{q}$. Then, $x (kq) = \frac{p(kq)}{q} = pk \in \mathbb{N}$, so $x \in A_{kq} \forall k \in \mathbb{N}$.
Now, given $m \in \mathbb{N}$,there is some $k$ such that $kq > m$. So, for all $m$, $x \in \bigcup_{l \geq m} A_l$, taking $l=kq$. Since the above applies for all $m$, it follows that $x \in \displaystyle\bigcap_{m=1}^\infty \bigcup_{l \geq m} A_l = \limsup A_m$. Since $x$ was a rational number, and we already showed that irrationals can't be part of $A_n$, it follows that $\limsup A_n = \mathbb{Q}$.
The other way, suppose that $x = \frac{p}{q} \in \liminf A_n =\displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$. This means that $x \in \bigcap_{k \geq n} A_k$ for some natural number $N$. This means, that $x(N+2)=\frac{p(N+2)}{q}$ and $x(N+1)=\frac{p(N+1)}{q}$ are both integers. By subtraction, we get that $\frac{p}{q}$ is also a natural number. But then, $x$ itself is a natural number. So every number in the $\liminf$ is a natural number.
On the other hand, if we have a natural number $q$, then $q \in A_n$ for all $n$, because taking $m= qn$, we see that $q = \frac{qn}{n}$, so $q \in A_n$. Since this applies for all $n$, certainly $q \in \bigcap_{n \geq 1} A_n$, and hence $q \in \displaystyle\bigcup_{n=1}^\infty \bigcap_{k \geq n} A_k$. Thus, all natural numbers are in the $\liminf$.
Thus, $\liminf A_n= \mathbb{N}$. I hope this was a clearer solution than the previous one.
Best Answer
Let us start from the line $x\in \bigcap_{k\geq n_0} A_k$. This implies that $x\in A_k$ for all $k\geq n_0$ and hence $x\in \bigcup_{k\geq n} A_k$ for all $n\geq 1$. Therefore also $x\in\limsup A_n$.