Solution: Let $(X,\mathcal{T} )$ be the indiscrete topological space. Recall that a set $A\subseteq X$ is said to be dense if $\bar{A} =X$, where $\bar{A} $ is the closure of $A$ which is defined as the intersection of all closed sets containing $A$. In this problem, we need to show that $\bar{A} = X$ holds for every nonempty subset $A$ of $X$. Since, $X$ is equipped with the indiscrete topology, only open sets are $\emptyset$ and $X$ so, only closed sets will be $\emptyset$ and $X$. Thus, \[ \bar{A} = X, \] as $X$ is the only closed set which contains $A$ and hence $A$ is dense in $X$.