For any topological space $X$, as the title explains, how many ways to construct a dense subspace of $X$? For example, we can construct a dense subspace which is the union of disjoint open subsets of $X$.
Added: If I may ask more, if $X$ is compact, do we have more ways to construct a dense subspace of $X$?
On
(Assuming you mean "how many" in the sense of cardinality.)
If $X = \mathbb{R}$ (or a subset of $\mathbb{R}$), there are $2^c$ dense subsets, where $c$ is the cardinality of the continuum.
Let $A$ be a countable dense subset of $X$ (for example, an enumeration of the rational points in $X$). Then $A \cup B$ is dense for any $B \subset X \setminus A$, so by varying $B$ you get $2^c$ different possibilities. On the other hand, there are "only" $2^c$ subsets of $X$, so the number of dense subsets can't be larger than this.
(You should be able to generalize this to many other toplogical spaces.)
Hint: Every compact metric space has countable base and is seperable, i.e it has a countable dense subset.
for $n\in \mathbb{N}$ the open sets $N_{1/n}(x)$ for $x\in X$ forms a open cover of $X$. since $X$ is compact, chhose a finite subcover $\{N_{1/n}(x_{n,1}),\dots,N_{1/n}(x_{n,k_n})\}$, note that for each $n$ the collection of $x$'s is different; therefore they must be labelled by both $n$ and a second parameter.
consider the countable base for $X$ $$\mathfrak{B}=\{N_{1/n}(x_{n,j}:n\in\mathbb{N}, 1\le j\le k_n\}$$ , now take any open set $U\subseteq X$ chose $\epsilon>0$ such that $N_{\epsilon}(x)\subseteq U$....enough?