Is a dense set always infinite? Learning about dense sets the classical example is that of $\mathbb Q$, the rationals, in $\mathbb R$.  The same interpretation is valid for irrationals in $\mathbb R$. I was wondering if a dense set needs to be infinite, because this is what intuition would suggest. Moreover, are dense sets always countably infinite?
 A: A dense set of the reals is always infinite. It need not be countable: For example, the reals are dense in themselves. 
And a dense set needs not be infinite either. For example, $\{1\}$ is a finite set that is dense in itself.
A: Every dense subset of $\mathbb R$ is infinite. A finite set has a maximum so it clearly cannot be dense. Dense subsets of $\mathbb R$ need not be countable; for example, the irrational numbers and the entire set $\mathbb R$ are both uncountable and dense. 
A: Certainly a set dense in the real line must be infinite: if $X$ is a finite subset of the real line, it has some maximum element $m$.  Then no point in $X$ is close to $m+1$ (otherwise $m$ is not the maximum, contradiction).  So $X$ cannot be dense. 
On the other hand, you mention that the irrational numbers are also dense, but these are already uncountable, so there’s no reason to think a dense set should be countable. (In fact, the more important thing is whether there is a countable dense set. This is called separability; see https://en.m.wikipedia.org/wiki/Separable_space).
A: In an infinite metric space any dense set must be infinite too: all finite sets are closed and so their closure ( the same finite set) cannot equal the whole space. One dense set is the whole space, so it need not be countable, unless the whole space is.
A: To answer the general question of "are dense sets always infinite": no, because certainly, if $X$ is a finite topological space, then $X$ is dense in itself.
For another example, if $X$ is any set with the indiscrete topology, then every nonempty subset of $X$ is dense.
For yet another example, let $X = \mathbb{R}$ with the topology determined by the Kuratowski closure operator
$$\operatorname{cl}(S) = \begin{cases} S, & 0 \notin S; \\ \mathbb{R}, & 0 \in S.\end{cases}$$
Then $\{ 0 \}$ is dense in $X$, yet $X$ is $T_0$.  (In fact, this example can easily be modified to give a $T_0$ topological space of any desired cardinality such that some single point is dense.)
On the other hand, in a $T_1$ topological space, every finite subset is closed.  So, if a finite subset of a $T_1$ topological space $X$ is dense, then $X$ itself must be finite.  Or, for the contrapositive, if $X$ is an infinite $T_1$ topological space, then every dense subset of $X$ is infinite.
