An example given in the textbook Foundation of topology by C.W. Patty is given below.
Let $X=\{(x,y)\in \mathbb R^2:x=0 \text{or } x=\frac{1}{n}$ for some $n \in \mathbb N$, and $0\leq y\leq 1\}$. I know that equivalence class $[(0,0)]=\{(x,y)\in \mathbb R^2|(x,y)$ and $(0,0)$ in the same connected component$\}$. Why $[(0,0)]$ is not an open set? Under what topology it is not open? Seeing $(X,\tau)$ as a subspace topology of usual topology on $\mathbb R^2$? $[(0,0)]$ can not be written as the intersection of $X$ with any open ball in $\mathbb R^2$. am I correct?
Note that any ball around zero in $\mathbb{R}^2$ will contain $(1/n,0)$ for some $n \in \mathbb{N}$. This means no ball will contain exactly $(0,0)$ without including other points in $X$. This means it cannot be expressed as the intersection of an open set in $\mathbb{R}^2$ with $X$ it is not open in the subspace topology. This becomes more transparent by drawing $X$.