The looped line topology (Willard #4D) I am having some trouble solving the following question in Willard's General Topology (p.36).

At each point $x$ of the real line other than the origin, the basic neighbourhoods of $x$ will be the usual open intervals centred at $x$. Basic neighbourhoods of the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$.
Q1. Verify that this gives a topology on the line.
Q2. Describe the closure operation in the resulting space.

For Q1, to show that the basic neighbourhoods give a topology on $\mathbb{R}$, I have to show that the following properties hold (from Theorem 4.5 in the textbook):
Edit: as pointed out in the comments, my description of the following theorem was inaccurate before. It has been updated - thank you to Brian M. Scott for pointing this out.
a) If $V \in \mathfrak{B}_x$ then $x \in V$.
b) If $V_1, V_2 \in \mathfrak{B}_x$ then $\exists V_3 \in \mathfrak{B}_x : V_3 \subseteq V_1 \cap V_2$.
c) If $V \in \mathfrak{B}_x$ then $\exists V_0 \in \mathfrak{B}_x$ such that $\forall y \in V_0, \exists W \in \mathfrak{B}_y : W \subseteq V$.
This is not hard to show now that I have the theorem copied correctly. However, I am still having problems with Q2.
How does one go about comprehensively describing the closure operation? Based on the above, the open sets are:

*

*The bounded, open sets $U \subseteq \mathbb{R}$ under the standard topology, where $0 \notin U$.

*The open sets $U \subseteq \mathbb{R}$ under the standard topology, for which $0 \in U$ and $(-\infty, a) \cup (b,\infty) \subseteq U$ for some $a<0, b>0$.

Obviously, the closed sets are the complements of open sets. But I am having trouble figuring out a concise way to describe the closure operation. Any further insight is appreciated.
 A: You’ve stated (c) incorrectly. In fact, it makes no sense as you’ve stated it: you can’t talk about the interior of a set until you have a topology, and the point of the first question is to show that you do have a base for a topology. Once you’ve done that, you can define the topology using (d) of Willard’s Theorem $4.5$. The correct statement of (c) is:

If $V\in\mathfrak{B}_x$, there is some $V_0\in\mathfrak{B}_x$ such that if $y\in V_0$, then there is some $W\in\mathfrak{B}_y$ such that $W\subseteq V$.

Verifying (a) is easy. If $x\ne 0$, members of $\mathfrak{B}_x$ have the form $(x-r,x+r)$ for some $r>0$, and clearly $x\in(x-r,x+r)$. And if $x\ne 0$, every $V\in\mathfrak{B}_x$ contains a subset of the form $(-\epsilon,\epsilon)$ for some $\epsilon>0$, and clearly $0\in(-\epsilon,\epsilon)\subseteq V$.
Verifying (b) is almost as easy. Suppose that $V_1,V_2\in\mathfrak{B}_x$. If $x\ne 0$ there are $r_1,r_2>0$ such that $V_1=(x-r_1,x+r_1)$ and $V_2=(x-r_2,x+r_2)$. Let $r=\min\{r_1,r_2\}$, and let $V_3=(x-r,x+r)$; then $V_3\in\mathfrak{B}_x$, and $V_3\subseteq V_1\cap V_2$. If $x=0$, there are $n_,n_2\in\Bbb Z^+$ and $\epsilon_1,\epsilon_2>0$ such that $V_i=(-\epsilon_i,\epsilon_i)\cup(\leftarrow,-n)\cup(n,\to)$ for $i=1,2$. Let $\epsilon=\min\{\epsilon_1,\epsilon_2\}$ and $n=\max\{n_1,n_2\}$, and let $V_3=(-\epsilon,\epsilon)\cup(\leftarrow,-n)\cup(n,\to)$; then $V_3\in\mathfrak{B}_0$, and $V_3\subseteq V_1\cap V_2$.
And because it will turn out that all of these basic nbhds are actually open in the topology that they generate, it’s also easy to verify (c). Let $V\in\mathfrak{B}_x$, and suppose first that $x\ne 0$, so that $V=(x-r,x+r)$ for some $r>0$. Let $s=\min\{r,|x|\}$, and let $V_0=(x-s,x+s)\in\mathfrak{B}_x$. If $y\in V_0$, let $t=\min\{(x+s)-y,y-(x-s)\}$; then $(y-t,y+t)\subseteq V_0$, and $(y-t,y+t)\in\mathfrak{B}_y$, since $y\ne 0$. (We needed $s$ in order to ensure that $0\notin V_0$.)
Now suppose that $x=0$, so that $V=(-\epsilon,\epsilon)\cup(\leftarrow,-n)\cup(n,\to)$ for some $\epsilon>0$ and $n\in\Bbb Z^+$. I’ll leave it to you verify that we can take $V_0=V$: show that for each $y\in V$ there is a $W\in\mathfrak{B}_y$ such that $W\subseteq V$.
The second question becomes very easy once you realize that this space is homeomorphic to an $\infty$ sign in the plane, e.g., to the union of the circles of radius $1$ with centres at $\langle -1,0\rangle$ and $\langle 1,0\rangle$. The topology is designed to bend the ends of the real line back around to approach $0$, so that $0$ becomes the centre point of the $\infty$.
See if you can finish it off with that much of a clue.
A: You haven't really addressed what to do when $y$ is also $0$.  In such a case, your parenthetical makes the false claim that we can take $W = (-\varepsilon, \varepsilon)$, but this is not an element of $\mathfrak{B}_y = \mathfrak{B}_0$.
Your handling of $x \neq 0$ is fine.
For $x = 0$, I might write...
For $x = 0$, let $V \in \mathfrak{B}_x = \mathfrak{B}_0$.  Then, there are $\varepsilon > 0$ and $n > 0$ such that $V = (-\infty, -n) \cup (-\varepsilon, \varepsilon) \cup (n, \infty)$.  Notice $V = V^\circ$, so let $y \in V$.
If $y \in (-\infty, -n) \in \mathfrak{B}_y$, we may take $W = (-\infty, -n) \subseteq V$.  A similar argument is used for $y \in (n,\infty) \in \mathfrak{B}_y$.
If $y \in (-\varepsilon, \varepsilon)$, either $y = 0$ or $y \neq 0$.  If $y = 0$, let $W = V \in \mathfrak{B}_x = \mathfrak{B}_y = \mathfrak{B}_0$.  If $y \neq 0$, take $W = (y/2, 2y) \cap (-\varepsilon, \varepsilon) \in \mathfrak{B}_y$.  Since
$$  -\varepsilon < -|y| < -|y/2| < |y/2| < |y| < \varepsilon  \text{,}  $$
this intersection is a nonempty open neighborhood of $y$.
