Let $P$ be a perfect $P\subset\Bbb R.$ Then there exists a family $\{P\_{\alpha}\subset P\colon \alpha<\mathfrak c\}$ of pairwise disjoint perfect subsets such that $$P=\bigcup\_{\alpha<\mathfrak c} P\_{\alpha}.$$ This is a well known result. The proof will rely on the following two theorems:

**Theorem** $1$. Let $X$ be a nonempty perfect Polish space. Then there is an embedding of $C$ (Cantor set) into $X.$

**Theorem** $2$. Let $C$ be a Cantor set. Then, $$C\cong C\times C=\bigcup\_{x\in C} \Big(\{x\}\times C\Big)$$

**Edit**: I thought it would be clear to get from the above Theorems but this is not the case even for me. Here what I got. Theorem 1 gives me, there is a map $f\colon C\to X$ establishing $C\cong f[C]\subset X.$ Let $P\subset X$ be a perfect$\subset X.$ Then it is polish space in itself. So, we can apply the Theorem 1. for a Cantor set $C'$ so there is map $g\colon C'\to P$ such that a $$\bigcup\_{x\in C'} \Big(\{x\}\times C'\Big)\cong g[C']\subset P$$ Now, this give us that $P$ contains continuum many pairwise disjoint perfect set. But How can we get equality? Any idea? In addition, is there any direct proof without using these Theorems?