Why are there exactly 2$^\omega$ perfect subsets of the real numbers? How can you proof that there are $2^{\omega}$ perfect subsets of the real numbers?

Let $C$ be the middle-thirds Cantor set, and for each $x\in\Bbb R$ let $C+x=\{y+x:y\in C\}$ be the translate of $C$ by $x$. Clearly each $C+x$ is perfect, and $C+x\ne C+y$ if $x,y\in\Bbb R$ with $x\ne y$, so $\{C+x:x\in\Bbb R\}$ is a family of $2^\omega$ perfect subsets of $\Bbb R$. This shows that there are at least $2^\omega$ perfect subsets of $\Bbb R$.
By definition a perfect set is closed, so to finish the argument it suffices to show that $\Bbb R$ has only $2^\omega$ closed subsets. Since there is an obvious bijection between the closed sets in $\Bbb R$ and the open sets, we need only show that there are just $2^\omega$ open sets in $\Bbb R$. $\Bbb R$ has a countable base $\mathscr{B}$ (e.g., the family of open intervals with rational endpoints). For each open set $U\subseteq\Bbb R$ there is a $\mathscr{B}_U\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{B}_U$. There are only $\omega^\omega=2^\omega$ subsets of $\mathscr{B}$, so there are at most $2^\omega$ open sets in $\Bbb R$ and hence at most $2^\omega$ perfect sets.
Thus, there are exactly $2^\omega$ perfect subsets of $\Bbb R$.
