Q:

Neighborhoods of $0$ in the co-countable topology on $\mathbb{R}$

Define $\tau = \emptyset \cup \{E \subset \mathbb{R} : E^c \:\: \text{is countable}\}$. I have shown that this is a topology on $\mathbb{R}$.  Now I want to show that the set of neighborhoods of $0$ does not have a countable base.
This is an old qualifier question.  I have not taken an actual topology course, so any help would be appreciated.

A:

To expand on Daniel's comment: note that
$$
\bigcap_{n \in \Bbb N} V_n = [\bigcup_{n \in \Bbb N} (V_n)^c]^c
$$
and that $\bigcup_{n \in \Bbb N} (V_n)^c$ is the countable union of countable sets, which is therefore countable.
Now take $\bigcap_{n \in \Bbb N} V_n$, and remove a single point.

