I recommend first grasping the intuitive idea behind $\displaystyle\bigcap_{k=1}\bigcup_{n\ge k}E_n$ and $\displaystyle\bigcup_{k=1}\bigcap_{n\ge k}E_n$. When you have a clear picture in your head, you'll be able to write a formal proof.

For example, $\displaystyle\bigcap_{n\ge k}E_n$ consists of elements belonging to all of $E_k$, $E_{k+1}$, ... . Eeach such element belongs to all but finitely many $E_n$'s. But elements of $\displaystyle\bigcap_{n\ge k+1}E_n$ and $\displaystyle\bigcap_{n\ge 2k}E_n$ also have this property. If you want to collect all elements that belong to every $E_n$ from some point, i.e., to form $\liminf E_n$, you need to take the union over $k$: $\displaystyle\bigcup_{k=1}\bigcap_{n\ge k}E_n$.

Similarly, $\displaystyle\bigcup_{n\ge k}E_n$ consists of elements that belong to some of $E_k$, $E_{k+1}$, ... ; that is, they occur in one of the sets after index $k$. If we take the intersection over $k$, we'll get the set of elements that belong to one of the sets starting from any point. I.e., they belong to one of the sets after $E_5$, they belong to one of the sets after $E_{100}$ and so on. That is, the resulting set has elements that belong to infinitely many $E_n$'s, i.e., $\limsup E_n$.