Does Every Subset of a Separable Topological Space have Countably Many Isolated Points?
No, there are many counterexamples. One of my favourites is Mrówka's $\Psi$ space, which I talked about in this answer, see also this blog post for much more info.
It's basically a countable open subset $D$ of isolated points that is dense in $X$ while $X\setminus D$ is uncountable and discrete as a subspace (so all its points are isolated within that set). The rational sequence topology on $\Bbb R$ is another instance of the same idea and also works as a more elementary counterexample.
It's indeed true for metric spaces in general. If a metric space is separable, it's second countable and thus hereditarily separable and hereditary Lindelöf and both of those last properties imply that all discrete (in themselves) subspaces are at most countable, which is what you were trying to show.
A space $X$ where a discrete subspace is at most countable is said to have countable spread, denoted by $s(X) = \aleph_0$. (Separable is countable density, $d(X)=\aleph_0$, second countable is called countable weight, $w(X)=\aleph_0$, and many other so-called cardinal invariants of spaces have been defined and studied, as well as their relationships. In these terms I have given counterexamples to the hypothesis $s(X) \le d(X)$ while in metric spaces $d(X)=hd(X)$ so there $s(X) \le d(X)$ does hold.)
A simple example is the Sorgenfrey plane, i.e., the plane $\mathbb R\times\mathbb R$ with the topology generated by the half-open rectangles $[a,b)\times[c,d)$. The set $\mathbb Q\times\mathbb Q$ of all rational points is a countable dense set, and the anti-diagonal $\{(x,-x):x\in\mathbb R\}$ is an uncountable discrete closed subset.
Another nice example is the compact Hausdorff space $\{0,1\}^\mathfrak c$, the product of continuum many two-point discrete spaces, which can be shown to be separable. The set of all points with a single nonzero coordinate is an uncountable discrete subset.
The Niemytzki plane (or Moore plane) is a counterexample.
- The family of all pairs of rational numbers is dense in the space, so it is separable.
- The $x$-axis $\{ (x,0) : x \in \mathbb{R} \}$ is an uncountable discrete subspace.
The only property I can think of at the moment which would imply that every subspace has countably many isolated points is hereditary separability, meaning that all subspaces are separable. (If $X$ has a subset $A$ with uncountably many isolated points, then $B \subseteq A \subseteq X$ consisting of the isolated points of $A$ would be an uncountable discrete subspace of $X$, which cannot be separable.) I am not certain at the moment if this is equivalent to having all subsets having countably many isolated points, but I would doubt it.