Consider the set $E = \left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots\right\} \cup \left\{1\frac{1}{2},1\frac{1}{3}, \cdots\right\}$. Observe that when we consider the space to be $\mathbb{R}$, $0$ and $1$ are the only limit points of the set $E$ but $\{0,1\}$ is not closed.
For ref., A set is closed in space $\mathcal{X}$ if it contains all its limit points.
Can anyone tell me with an example what this post is about?
According to your definition of closed set, a set $A$ is not closed if there exists a limit point $x$ of $A$ such that $x\notin A$.
Now, what are the limit points of $\{0,1\}$? Can you find one that does not belong to $\{0,1\}$?
Suppose $x$ is a limit point of $A$ in the metric space $(X,d)$. Then there exists $x_0\in A$ such that $x\ne x_0$ and $d(x,x_0)<1$.
Suppose you have found $x_0,x_1,\dots,x_n\in A$, $x_k\ne x$, pairwise distinct. Now, by assumption, there exists $x_{n+1}\in A$, $x_{n+1}\ne x$ with $d(x_{n+1},x)<d(x_n,x)/2$. In particular, the points $x_0,x_1,\dots,x_n,x_{n+1}$ belong to $A$ and are pairwise distinct.
The recursion doesn't end, so you have a sequence $(x_n)$ of pairwise distinct points of $A$. Hence $A$ is infinite.
