Can we have an uncountable number of isolated points? Is this possible? I've been trying to think of an example or defend why not, and I'm struggling in both directions.
 A: As others have noted, while the standard topology of $\mathbb R$ (or $\mathbb R^n$) admits no uncountable discrete subset, other topological spaces (such as, trivially, $\mathbb R$ with the discrete topology) do contain such subsets.
This is even possible for a locally Euclidean space, provided that it's simply "big enough".  A natural example is the "long line" obtained by gluing together an uncountable number of (specifically, $\omega_1$) copies of the unit interval.  The long line is locally homeomorphic to $\mathbb R$, and also equinumerous with it, yet in a certain topological sense much "longer" — for example, it is not metrizable or second countable.  Also, clearly, picking, say, the midpoint of each of the unit intervals yields an uncountable discrete subset of the long line, whereas the real line has none.
In fact, Brian M. Scott's argument naturally generalizes to any second countable space, showing that no such space can have an uncountable discrete subset.
(The converse, however, does not hold: there are spaces that are not second countable, but which do not have an uncountable discrete subset.  A simple example is the quotient space $\mathbb R / \mathbb Z$, i.e. the space obtained by taking the real line and "gluing together" all the integers.  This space is neither second nor even first countable (because the "central point" formed by the glued-together integers has no countable neighborhood basis), yet any uncountable discrete subset of it (which we may assume does not include the central point, since we can remove it without affecting countability) would also have to be discrete as a subset of $\mathbb R$.)
A: Since you use the real-analysis tag, I will assume that you’re talking about $\Bbb R$ with the usual topology. If so, the answer is no: every discrete subset of $\Bbb R$ is countable. 
To see this, note first that the set $\mathscr{B}$ of open intervals with rational endpoints is a countable base for the topology. Now suppose that $D\subseteq\Bbb R$ is discrete. Then for each $x\in D$ there is a $B_x\in\mathscr{B}$ such that $B_x\cap D=\{x\}$. $\mathscr{B}$ is countable, so if $D$ were uncountable, there would have to be distinct points $x,y\in D$ such that $B_x=B_y$, which is absurd: that would mean that 
$$\{x\}=B_x\cap D=B_y\cap D=\{y\}\;,$$
yet $x\ne y$.
If you are asking the question in general, however, rather than about $\Bbb R$ with the usual topology, then the answer is yes; Ross Millikan has given a simple example.
A: In $\mathbb{R}$ with the usual topology, the answer is no - but the reason is a little more complicated than "there is a countable dense set." Consider the topology on $\mathbb{R}$ generated by $\{X\subseteq\mathbb{R}: 0\in X\}$, that is, a set is open if it is empty or contains $0$. Then $\{0\}$ is dense, but if we let $A=\mathbb{R}-\{0\}$, then every element of $A$ is an isolated point in $A$.
A: You did not specify the space you are working in, nor the topology.  The simplest example I can think of is $\Bbb R$ with the discrete topology. All points are isolated.
A: In the discrete topology, every singleton $\{ x\}$ is open and hence is a neighbourhood with only $x$: thus $x$ is isolated.
A: I'll just expand on one of the comments. The union of open disjoint intervals is countable. This is because we may find a one-to-one correspondence between such a union and a subset of rationals. 
Let $A$ have infinite isolated points. Let $x,y\in A$ be isolated with $x\neq y,$ we may find a neighborhood of $x$ that does not contain $y$ and vice-versa. There is also a neighborhood for each that do not contain an element of $A$ because $x$ and $y$ are isolated points. Take the min of those lengths to get two disjoint neighborhoods, one centered at $x$ and the other at $y$. 
Let $\Delta$ be the set of all isolated points of $A$. Using the process given above, there are neighborhoods $V_{\epsilon_{x}}(x),x\in \Delta$ that are pairwise disjoint. Pick one and only one rational in each $V_{\epsilon_{x}}(x)$. Then 
$\displaystyle \bigcup_{x\in \Delta}V_{\epsilon_{x}}(x)$ is equivalent to a subset of $\mathbb{Q}$.
A: No, each neighborhood around each point is disjoint and contains a rational not contained in any other point's neighborhood. 
