Consider the set $A = \{\frac 1 n : n \in \Bbb{Z}^+\} = \{1, 1/2, 1/3, 1/4, ...\}$. This has exactly one limit point, namely $0$.
Next consider the set $\{1 + \frac{1}{n} : n \in \Bbb{Z}^+\}$. This has exactly one limit point, namely $1$. Now consider the union of these two sets: Do you see why it has exactly two limit points? Do you see how to extend this to having $3$, or $n$, limit points?
To show that $A$ actually has $0$ as a limit point, just note that if given $\epsilon > 0$, there exists an $n$ for which $\frac{1}{n} < \epsilon$. So every neighborhood of $0$ intersects $A$ at a point.
Now suppose that $\alpha$ is a limit point of $A$. If $\alpha < 0$, set $\epsilon = \frac{-\alpha}{2}$, and note that $(\alpha - \epsilon, \alpha + \epsilon) \cap A = \emptyset$, a contradiction. Likewise, $\alpha > 1$ leads to a contradiction. If $\alpha = 1$, set $\epsilon = \frac{1}{3}$.
Finally, if $0 < \alpha < 1$, let $n$ be the least integer satisfying $\frac{1}{n} < \alpha$ and choose $\epsilon = \frac{1}{2} (\alpha - \frac 1 n)$.
Best Answer
Let us assume that $A$ has only finitely many limit point, $a_1, \dotsc, a_n$.
For every integer $k$, we form a set $X_k = \bigcup\limits_{i = 1}^n[a_i - 1/k, a_i + 1/k]$.
It is a simple exercise to check that the intersection $\bigcap\limits_{k = 1}^\infty X_k$ is just the set $\{a_1, \dotsc, a_n\}$.
In particular, the set $B = A \backslash \bigcap\limits_{k = 1}^\infty X_k$ is still uncountable.
But we can rewrite: $$ B = A \cap \left(\bigcap_{k = 1}^\infty X_k\right)^c = A \cap \left(\bigcup_{k = 1}^\infty X_k^c\right) = \bigcup_{k = 1}^\infty (A \cap X_k^c) = \bigcup_{k = 1}^\infty (A\backslash X_k).$$
Let $B_k$ denote the set $A\backslash X_k$. Since $B$ is the countable union of all $B_k$, the uncountability of $B$ implies that at least one $B_k$ is uncountable.
However, if some $B_k$ is uncountable, then by applying part a) of the question to $B_k$, we know that it must have a limit point. This limit point cannot be any of the $a_i$, as an interval around each of them is taken away.
Furthermore, since $B_k$ is contained in $A$, a limit point of $B_k$ is also a limit point of $A$.
This contradicts the original hypothesis that the $a_i$'s are all the limit points of $A$.