(a) Let $\{U_n\}$ be a countable basis at $x$ . Then for each $y\neq x$ there is a neighborhood $U$ of $x$ such that $U$ does not contain $y$ . Therefore, there is $U_k$ that does not contain $y$ . Thus, $\cap_n U_n=\{x\}$ .(b) Given any $x$ and $y\neq x$ there must be a neighborhood of $x$ that does not contain $y$ , therefore, the space is a $T_1$ -space and every one-point set is closed. We can start with some first-countable $T_1$ -space. By (a), every one-point set in such a space will be a $G_\delta$ set. Then, if we take a finer topology, it will still have the property, however, we may loose the first-countability if the space is large enough. For example, if we take $R^\omega$ then it is a metric space in the product topology. Therefore, every one-point set is the intersection of the countable collection of balls of radius $1/n$ centered at the point. Now, if we switch to the box topology, then it seems not to satisfy the first-countability axiom (take a countable collection $U_n$ of neighborhoods of a point and construct a neighborhood by taking the product of open proper subsets of $U_{nn}$ ).Also, another example would be any countable $T_1$ -space which is not first-countable. Such a space will also be separable and Lindelöf but not second-countable. One example of such a space I found by solving Exercise 17.