I have this topology on $\mathbb{R}^2$ $$\sigma=\{\emptyset\}\cup \big\{\bigcup_{r\in A}\Omega_r, A\subseteq \mathbb{R}\setminus\{0\} \big\},$$
where $\Omega_r= \big\{(x,y)\in\mathbb{R}^2, (x+2)^2+(y-2)^2<r^2 \big\}$.
The question is to find all the possible limits of $\big(\frac1n,0 \big)$.
I begin by finding the open set that contains the sequence: $$ \big(\frac1n,0 \big)\in \Omega_r\Longleftrightarrow \big(\frac1n+2 \big)^2+4<r^2\Rightarrow \frac1n<\sqrt{r^2-4}-2\Rightarrow n>\frac{1}{\sqrt{r^2-4}-2} $$ then $$ \forall r>\sqrt{8}, \exists n_0= E\left[\frac{1}{\sqrt{r^2-4}-2}\right]+1, \forall n\in \mathbb{N}, n\geq n_0\Rightarrow \big(\frac1n,0 \big)\in \Omega_r.$$
What should the conclusion look like?
To straighten the twisted definition of the topology,
R^2 is given the base B = { $U_r$ : r > 0 } where
$U_r$ = { (x,y) : $(x + 2)^2 + (y - 2)^2$ < $r^2$ },
an open disk of radius r, centered at (-2,2).
Let t = 2$\sqrt2$, the distance from the center of $U_r$
to (0,0) and C = { (x,y) : $(x + 2)^2 + (y - 2)^2$ = $t^2$ }.
As the space is not Hausdorff, limits of
the sequence (1/n,0) don't have to be unique.
Indeed, every point of C, (0,0) included, is a limit of the
sequence (1/n,0). An exericise I leave for the reader.
In fact, every point not in $U_t$ is a limit of the sequence (1/n,0).