Another way for partition of perfect set 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?
 A: Definitions. $\mathbb R$ is the real line. A subset of $\mathbb R$ is an interval if it's connected and has more than one point, a perfect set if it's a nonempty closed set with no isolated points, a Cantor set if it's homeomorphic to the standard Cantor set $C$.
Lemma. Every perfect set $X\subseteq\mathbb R$ can be partitioned into subsets each of which is either an interval or a Cantor set.
Proof. Let $Y$ be the union of all sets of the form $U\cap X$ where $U$ is an open subset of $\mathbb R$ and $U\cap X$ is connected, and let $Z=X\setminus Y$. (Either $Y$ or $Z$ may be empty.)
Note that $Y$ can be partitioned into intervals; namely, each component of $Y$ is an interval.
Note that $Z$ is a totally disconnected closed set with no isolated points. If $Z$ is nonempty and compact, then $Z$ is a Cantor set; if $Z$ is not compact, $Z$ can be partitioned into countably many Cantor sets. In either case, $Z$ can be partitioned into Cantor sets.
Theorem. Every perfect set $X\subseteq\mathbb R$ can be partitioned into $\mathfrak c$ Cantor sets.
Proof. By the lemma $X$ can be partitioned, if not into Cantor sets, then at least into Cantor sets and intervals. However, by Theorem 1.14 (and the first sentence of the proof) of Paul Bankston and Richard J. McGovern, Topological partitions, General Topology and its Applications 10 (1979), 215–229 (pdf), each interval can be partitioned into Cantor sets. Hence $X$ can be partitioned into Cantor sets. Since a Cantor set can be partitioned into $\mathfrak c$ Cantor sets, it follows that $X$ can be partitioned into $\mathfrak c$ Cantor sets.
A: I think the fact that I mentioned in OP wouldn't behold for a perfect set in ZFC by the following exercise, see fundamentals of real analysis by James Foran, exercise $13$ page $127.$
Exercise: Prove that the continuum hypothesis implies that each perfect set subset of the real line can be written as the union of $\omega_1$ pairwise disjoint non-empty perfect sets.
If OP is correct for every perfect set so the above exercise has no meaning since $\omega_1=\mathfrak c.$
Is that correct? Please share your thought. Thank you all.
