We wish to show that $A = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$ is the set of elements that lie in infinitely many $E_{k}$.
To see this, we first note what this notation is saying. We first take the union $\bigcup _{k=1}^{\infty} E_{k}$, then we intersect it with the union $\bigcup_{k=2}^{\infty} E_{k}$, which is equivalent to throwing away $E_{1}$. Then, we intersect with $\bigcup_{k=2}^{\infty} E_{k}$, which means we are throwing away $E_{2}$. If we keep on throwing away $E_{n}$, the only elements that remain are those that are in $E_{k}$ for arbitrarily large $k$, which is just the set of elements that lie in infinitely many $E_{k}$. With this intuition, we attempt a formal proof:
Let $x$ lie in infinitely many $E_{k}$. Then, for every $n$, there $j > n$ such that $x \in E_{j}$. Thus, $x \in \bigcup_{k=n}^{\infty}$ for every $n$, and hence it is in the intersection $\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$. For the converse, we do a proof by contrapositive. Suppose that $x$ does not lie in infinitely many $E_{k}$. Then there is some largest $E_{K}$ in which $x$ lies. Hence, for $n > K$, we have $x \notin \bigcup_{k=n}^{\infty} E_{k}$. Thus $x$ will not be in the intersection $\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} E_{k}$, as it is not in every $\bigcup_{k=n}^{\infty} E_{k}$.