Any uncountable subset $A$ of $\mathbb{R}$ has uncountably many limit points.
This is true in any separable metric space. Take $B_n$ a countable open base for the topology (for the reals we can take all open intervals with rational endpoints), and suppose $A$ is uncountable and consider points not in $A'$:
For every $x \in A\setminus A'$, pick $B_n(x)$ such that $B_n(x) \cap A \subset \{x\}$, this can be done as $x$ is not a limit point of $A$.
Clearly $x \neq y$ implies $n(x) \neq n(y)$.
So we have an injection from the set $A \setminus A'$ into the countable set of indices $\mathbb{N}$.
So there are only countably many points of $A$ that are not limit points of $A$, and as $A$ is uncountable, $A'$ must be uncountable.
So indeed a set with at most countably many limit points is at most countable as well.