We have basis of topology $\tau = \{\{-r,r\}: r>0\} \cup \mathbb{R}$. I want to find all sequences, which convergence to $x_1 = 0$ and $x_2= \frac{1}{3}$.
The neighbourhood of $x_1 = 0$ is $\mathbb{R}$. So it can be all sequences, which equal $0$?
The neighbourhood of $x_1 = \frac{1}{3}$ is $\{-\frac{1}{3}, \frac{1}{3}\}$. So, here its all sequences, which belong to $\{-\frac{1}{3}, \frac{1}{3}\}$?
Recall that a sequence $(x_n)_n$ in a topology space $(X,\tau)$ converges to a point $x \in X$, if for each neighbourhood $U$ of $x$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ we have that $x_n \in U$.
For $x_1=0$, you are right, there is a unique neighbourhood, i. e. $\mathbb{R}$. This implies that any sequence is convergent to $x_1=0$.
For $x_1=1/3$, you are almost correct. There are several neighbourhoods of $x_1$, but all such neighbourhoods contain the neighbourhood $\{-1/3,1/3\}$. It follows that a sequence is convergent to $x_2=1/3$ if and only if it is eventually contained in $\{-1/3,1/3\}$.
Note that $(X,\tau)$ is not a Hausdorff space and the limit of a sequence is not necessarily unique.