I think I finally managed to prove this exercise from Kelley's general topology on page 77.
I would appreciate feedback on it.
Let $X$ be the set of all pairs of $\Bbb N_0$ with the topology described as follows:
For each point $(m,n)$ other than $(0,0)$ the set $\{(m,n)\}$ is open.
A set $U$ is a neighbourhood of $(0,0)$ iff for all except a finite number of integers $m$ the set $\{n:(m,n)\notin U\} $ is finite.
(Visualizing $X$ in the Euclidean plane, a neighbourhood of $(0,0)$ contains all but a finite number of the members of all but a finite number of columns.)
(b) Each point of $X$ is the intersection of a countable family of closed neghbourhoods
Proof
Since any point other than $(0,0)$ is open, complement of the union of all open points is $(0,0)$. Hence it is closed.
So let $(a,b)$ be any point in $X$, let $\mathcal U$ be a countable family of neighbourhoods of $(0,0)$ s.t:
- Every member of $\mathcal U$ contains $(a,b)$;
- each member excludes finite amount of columns and finite points from column of $(a,b)$ which no other member excludes.
Then $\cap^\infty \{U:\, U\in\mathcal U\} = (a,b).$
It has a few problems. I think that you have the right idea for showing that $\{\langle 0,0\rangle\}$ is the intersection of a countable family of closed nbhds of $\langle 0,0\rangle$, but you’ve not stated it very clearly. For convenience let $D=X\setminus\{\langle 0,0\rangle\}$. For each $\langle m,n\rangle\in D$ let $$U(m,n)=X\setminus\{\langle m,n\rangle\}\;;$$ $U(m,n)$ is a nbhd of $\langle 0,0\rangle$, and since the point $\{\langle m,n\rangle\}$ is open, $U(m,n)$ is closed set. Clearly the family $\{U(m,n):\langle m,n\rangle\in D\}$ is countable, and
$$\begin{align*} &\bigcap\big\{U(m,n):\langle m,n\rangle\in X\setminus\{\langle 0,0\rangle\}\big\}\\ &\quad=\bigcap\big\{X\setminus\{\langle m,n\rangle\}:\langle m,n\rangle\in X\setminus\{\langle 0,0\rangle\}\big\}\\ &\quad=X\setminus\bigcup\big\{\{\langle m,n\rangle\}:\langle m,n\rangle\in X\setminus\{\langle 0,0\rangle\}\big\}\\ &\quad=X\setminus(X\setminus\{\langle 0,0\rangle\})\\ &\quad=\{\langle 0,0\rangle\}\;, \end{align*}$$
so $\{\langle 0,0\rangle\}$ is the intersection of a countable family of closed nbhds of $\langle 0,0\rangle$.
Your argument for the other points, however, definitely does not work: the members of your family $\mathscr{U}$ are all nbhds of $\langle 0,0\rangle$, so their intersection contains $\langle 0,0\rangle$ and therefore cannot be $\{\langle a,b\rangle\}$.
In fact if $\langle m,n\rangle\in D$, all you need to do is let $V=\{\langle m,n\rangle\}$ and let $\mathscr{V}=\{V\}$. Certainly $V$ is a nbhd of $\langle m,n\rangle$, $\mathscr{V}$ is countable, and $\bigcap\mathscr{V}=\{\langle m,n\rangle\}$, so it only remains to verify that $V$ is a closed nbhd of $\langle m,n\rangle$. But $V=X\setminus U(m,n)$, and we already observed that $U(m,n)$ is open nbhd, so $V$ is closed, and we’re done.