Assume that the only dense subset is $X$ itself. What can you say something about the topology, that is, the family of open sets? Let $(X, d)$ be a metric space. Assume that the only dense subset is $X$ itself. What can you say something about the topology, that is, the family of open sets?
My Try: Since the only dense subset of $X$ is $X$ itself, so every subset of $X$ is a closed set.

What does it mean for a subset $A$ of $X$ to be dense?  It means for each $x \in X$ and every open $U \subseteq X$ with $x \in U$, $U \cap A \neq \emptyset$.
So if $X$ is the only dense subset of itself, that means for each $x \in X$, $X - \{ x \}$ is not dense in $X$.  That means, in particular, that there is some $y \in X$ and some open $U \subseteq X$ such that $y \in U$ and $U \cap (X - \{ x \}) = \emptyset$.  But if this intersection is empty, then it must be that $y = x$ (why?).  It also must be that $U = \{ x \}$ (why?), and so $\{x \}$ is open.  But then that means that singletons are open.
But if $\{ x \}$ is open for each $x$, then it is easy to see that the topology on $X$ is actually the discrete topology (why?), i.e., the set of all subsets of $X$, which is the finest topology possible on $X$.
