Can a dense set contain isolated points? I was interested in this question, can a dense set contain an isolated point, because I was reading into the lexicographic order topology on the unit square. 
I read in here that:

$S$ is not separable, since the set of all points of the form $(x,\frac{1}{2})$ is discrete but is uncountable. 

I do understand why $T=\{ (x,\frac{1}{2} ) \,|\,\,\, 0\leq x \leq 1 \}$ is discrete, but I don't see how $T$ being uncountable assists us in showing $S$ is not separable? I thought maybe this is because a dense set cannot contain any isolated points? But I'm not sure on that. If anyone can clarify here, it'll be great!
 A: That explanation is very badly written. It isn’t the fact that the set is discrete that shows that $S$ is not separable: it’s the fact that the points of the set can be enclosed in pairwise disjoint open intervals, e.g., the intervals $\big(\langle x,0\rangle,\langle x,1\rangle\big)$ for $x\in[0,1]$. If $S$ had a countable dense subset $D$, each of these uncountably many intervals would have to contain a different member of $D$ which is obviously impossible.
There are many separable spaces with uncountable discrete subsets, even uncountable closed discrete subsets. One is the tangent disk space, also known as the Moore or Niemytzki plane.
As for your original question: if a space has isolated points, they must necessarily be part of any dense subset. This is clear from the fact that if $p$ is isolated, $\{p\}$ is an open set, and the only point that it contains is $p$.
A: Even if $X$ has no isolated points, it is possible that a dense set $S$ has isolated points. For example if $X$ is indiscrete, then any one-point set is dense and this point is isolated. Or if $X=A\times B$ where $A$ is discrete and $B$ is indiscrete (but $|B|>1$), the the graph of any function $A\to B$  is dense in $X$ and is a discrete space.
