Topological restrictions on cardinality From what I know, a Polish (completely metrizable separable) space has a cardinality at most of $\mathbb R$. Completeness assumption can be omitted here, because a completion of a metrizable separable space is Polish. On the other hand, without separability the cardinality of the space can be greater than that of $\mathbb R$ - we just can endow any set with a discrete topology.
My question is the following: can a separable topological space has the cardinality greater than that of $\mathbb R$? 
 A: Yes: $\beta\omega$ is a separable space of cardinality $2^{2^\omega}=2^\mathfrak{c}$. There’s a complete proof in this answer to an earlier question.
Assuming that the separable space $X$ is $T_2$, this is an upper bound on its cardinality. Let $D$ be a countable dense subset. For each $x\in X$ let $\mathscr{D}(x)=\{D\cap V:V\text{ is an open nbhd of }x\}$. Then $\mathscr{D}(x)$ is a family of subsets of $D$, so $|\mathscr{D}(x)|\le 2^\omega=\mathfrak{c}$. If $x,y\in X$ and $x\ne y$, disjoint nbhds of $x$ and $y$ have disjoint traces on $D$, so $\mathscr{D}(x)\ne\mathscr{D}(y)$. Thus, $$|X|=|\{\mathscr{D}(x):x\in X\}|\le|\wp(\wp(D))|=2^{2^\omega}=2^{\mathfrak c}\;.$$
Added: Originally I had written that $T_1$ separation was sufficient, but as t.b. points out, this is false: the cofinite topology on a set of any cardinality is a compact, separable, $T_1$ topology.
A: It depends if you require separation axioms to hold (regularity, Hausdorff, etc.)
If you simply require separability, let $X$ be a set as large as you would like it to be, along with the trivial topology.
Fix any $x\in X$. Now for every $y\in X$, and an open environment $U$ of $y$ we have that $x\in U$. Therefore $\operatorname{cl}(\{x\})=X$.
Similarly you can consider the topology of all sets $U$ such that $U=\varnothing$ or $x\in U$. In this topology $\{x\}$ is also dense.
In both topologies we have a dense singleton.
