From 'baby' Rudin.
I've seen that a set is closed iff it contains all of its limit points. In Rudin, $(d)$ says if every limit point of E is a point of E, then $E$ is closed. He also says $(h)$: $E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$.
But Closed $\implies$ contains all of its limit points. So, is every closed set a perfect set?
On
No finite set is perfect but every finite set is closed; a finite set has no limit points and thus all of its limit points (all zero of them) belong to it, so it's closed.
On
No. Consider the set $[0,1] \cup \{2\} \subseteq \mathbb{R}$ with the usual topology. By the definition of limit point, if $x_0$ is a limit point of the set $S \subseteq \mathbb{R}$, $$ \forall r > 0, \exists x \in \mathbb{R}\mid x \neq x_0, |x-x_0| < r $$ Clearly the point $2$ does not satisfy this condition(taking $r = 1/2$), and thus is not a limit point.
Closed means all limit points are in $E$. But that doesn't mean all points in $E$ are limit points. Any closed set with a point that is not a limit point will not be perfect.
The easiest counter example is a set with a single point. That set is closed but its one point isn't a limit point.
Less trivial and less contrived is $D = \{a + 1/n| a \in \mathbb Z,n \in \mathbb N\} $. Every integer is a limit point. No other point is a limit point. All integers are in $D$ (because $a + 1/1$ is an integer) so $D$ is closed. But for all $n > 1$ then $a + 1/n $ is in D but is not a limit point. So $D$ is not perfect.