[proofplan] We use the open-set characterization of density: a subset is dense exactly when it intersects every nonempty open subset. In the [indiscrete topology](/page/Indiscrete%20Topology) $\tau=\{\varnothing,X\}$, the only nonempty open subset is $X$ itself. Thus density is equivalent to requiring $A\cap X\neq\varnothing$, which is exactly the condition $A\neq\varnothing$ because $A\subset X$. [/proofplan] [step:Identify the only nonempty open subset of the indiscrete space] The open subsets of $(X,\tau)$ are precisely the elements of $\tau=\{\varnothing,X\}$. Since $X$ is nonempty, $\varnothing$ is not a nonempty open subset and $X$ is a nonempty open subset. Hence the only nonempty open subset of $(X,\tau)$ is $X$. [guided] The topology is given explicitly as $\tau=\{\varnothing,X\}$. Therefore a subset $U\subset X$ is open in $(X,\tau)$ exactly when $U=\varnothing$ or $U=X$. The hypothesis that $X$ is nonempty is used here. It ensures that $X$ itself is a nonempty open subset. Since $\varnothing$ is empty, it is not a nonempty open subset. Consequently, among all open subsets of $(X,\tau)$, the only one that is nonempty is $X$. [/guided] [/step] [step:Show that every nonempty subset is dense] Assume $A\subset X$ and $A\neq\varnothing$. By the previous step, the only nonempty open subset of $(X,\tau)$ is $X$. Since $A\subset X$, we have $A\cap X=A$, and therefore $A\cap X\neq\varnothing$. Thus $A$ intersects every nonempty open subset of $(X,\tau)$. By the open-set characterization of dense subsets, $A$ is dense in $(X,\tau)$. [/step] [step:Show that every dense subset is nonempty] Assume $A\subset X$ is dense in $(X,\tau)$. Since $X$ is a nonempty open subset of $(X,\tau)$, the open-set characterization of density gives $A\cap X\neq\varnothing$. Because $A\subset X$, we have $A\cap X=A$. Hence $A\neq\varnothing$. [/step] [step:Conclude the equivalence] We have proved that every nonempty subset $A\subset X$ is dense in $(X,\tau)$, and that every [dense subset](/page/Dense%20Subset) $A\subset X$ is nonempty. Therefore, for every subset $A\subset X$, the subset $A$ is dense in $(X,\tau)$ if and only if $A\neq\varnothing$. [/step]