No perfect non-empty set in $\mathscr{P}(\Bbb{R})$ can be strong measure zero. Here is what to be proven:

No perfect non-empty set in $\mathscr{P}(\Bbb{R})$ has strong measure zero.

The textbook (Teoría de la Medida, Jaime San Martin Aristegui, section 1.6) suggests the following approach:

*

*Suppose that $f\colon [0,1]\to \Bbb{R}$ is a continuous function and $A\subset [0,1]$ a strong measure $0$ set. Prove that $f(A)$ is also a strong measure $0$ set.

*Show that every perfect set $A\subset [0,1]$ (not empty) contains a set $F$ homeomorphic to the Cantor set and therefore $A$ can't be a strong measure $0$ set.

I have already proven 1 but have no idea how to prove 2; in fact, I don't think it's true (intuitively, I'm not saying that the text book is wrong). Any hints or solutions are more than welcomed.

The solution to the first point is available at this post,  If $A\subset [0,1]$ is strong measure zero, then $f(A)$ is strong measure zero where $f\colon [0,1]\to \Bbb{R}$ is continuous.
 A: Suppose we have a perfect nonempty set $A\subseteq [0,1]$. Call an interval $[a,b]$ proper if $a\not=b$.

*

*$A$ contains a proper interval.
Since a proper interval is homeomorphic to $[0,1]$, which contains the Cantor set, $A$ contains a set that is homeomorphic to the Cantor set.


*$A$ does not contain any proper interval. We will prove that $A$ is homeomorphic to the Cantor set. 
Recall that the Cantor set $\mathcal {C}$ is created by iteratively deleting the open middle-third part from a set of closed intervals, starting from $[0,1]$. Let us show that $A$ can be obtained similarly by iteratively deleting the open "middle-third part" from a set of closed intervals, starting from $[A_0,A_1]$, where $A_0=\inf\,(A)$ and $A_1=\sup\,(A)$. Since $A\subseteq[0,1]$ is nonempty and closed, $A_0$ and $A_1$ are in $A$ and $A\subseteq[A_0, A_1]$. Since a nonempty space that has no isolated points contains more than $1$ point, $[A_0, A_1]$ is a proper interval.
Middle-third part with respect to $A$
Call a proper interval $[a,b]$ an  $A$-interval if $a, b\in A$ and $(a-\epsilon,a)\cap A=(b,b+\epsilon)\cap A=\emptyset$ for some $\epsilon>0$.
Let $[a,b]$ be an $A$-interval. Since $A$ does not contain $[a+\frac{b-a}3,b-\frac{b-a}3]$, there is a point $m$ in that interval that is not in $A$. Let $m_-=\sup\,\{x\in A \mid x\lt m\}$ and $m_+=\,\inf \{x\in A \mid x\gt m\}$. Since $A$ is closed, $m_-$ and $m_+$ are in $A$. $\,m\in(m_-,m_+)$. $\,(m_-, m_+)\cap A=\emptyset$. So, $(m_-, m_+)$ is the maximal open interval containing $m$ that is disjoint from $A$.
Call the open interval $(m_-,m_+)$ the middle-third part of $[a,b]$ w.r.t. $A$. There might be many such intervals; any one will do. 
Since $a\in A$  is not an isolated point, $[a,m_-]$ is a proper interval and hence an $A$-interval. So is $[m_+,b]$.
Construction of space $\mathscr A$
For iteration $0$, there is $1=2^0$ $A$-interval $[A_0, A_1]$.
Let $(A_{\frac13}, A_{\frac23})$ be the middle-third part of $[A_0, A_1]$ w.r.t. $A$. Delete $(A_{\frac13}, A_{\frac23})$ from $[A_0, A_1]$, leaving two $A$-intervals: $[A_0, A_\frac13]$  and $[A_\frac23, A_1]$. This is iteration $1$, with $2^1$ disjoint $A$-intervals present.
Next, let $(A_\frac19, A_\frac29)$ be the middle-third part of $[A_0, A_\frac13]$ w.r.t. $A$ and let $(A_\frac79, A_\frac89)$ be the middle-third part of $[A_\frac23, A_1]$ w.r.t. $A$. Delete these two middle-third parts, leaving four intervals: $[A_0,A_\frac19]\cup [A_\frac29,A_\frac13]\cup [A_\frac23,A_\frac79]\cup[A_\frac89, A_1]$. This is iteration $2$, with $ 2^2$ disjoint $A$-intervals present.
Repeating this process infinitely. Note that the endpoints of all intervals at each iteration are in $A$.
Denote the set of points in $[A_0, A_1]$ that are never deleted by $\mathscr A$. Note that $\mathscr A$ is also the intersection of $A_n$ for all $n\ge0$, where $A_n$ is the union of all $2^n$ disjoint $A$-intervals at iteration $n$. 
Show $A=\mathscr A$

*

*Since no point in $A$ is ever deleted, $A\subseteq\mathscr A$.

*At iteration $n$, the length of each interval is $\le(2/3)^n$.
Suppose $x\in \mathscr A$.
At iteration $n$, there is a interval with length $\le(2/3)^n$ that contains $x$. Since both endpoints of that interval are in $A$, there is a point in $A$ that is $\le(2/3)^n$ away from $x$. So $x$ is a limit point of $A$.
Since $A$ is closed, we must have $x\in A$.

Hence, $A=\mathscr A$.
Homeomorphism
Let $h_{0,0}:(\frac13, \frac23)\to(A_\frac13, A_\frac23)$ be the linear map that maps  $\frac13$ to $A_\frac13$ and $\frac23$ to $A_\frac23$ restricted to $(\frac13, \frac23)$.
Let $h_{0,1}:(\frac19, \frac29)\to(A_\frac19, A_\frac29)$ be the linear map that maps $\frac19$ to $A_\frac19$ and $\frac29$ to $A_\frac29$ restricted to $(\frac19, \frac29)$. Let $h_{1,1}:(\frac79, \frac89)\to(A_\frac79, A_\frac89)$ be the linear map that maps $\frac79$ to $A_\frac79$ and $\frac89$ to $A_\frac89$ restricted to $(\frac79, \frac89)$.
Repeating the construction of $h{*,*}$ infinitely, each time mapping linearly each deleted interval during the construction of the Cantor set to the corresponding deleted interval during the construction of $\mathscr A$.
It can be seen that combining all $h_{*,*}$, we have defined a uniformly-continuous strictly-increasing map on an open set of $[0,1]$ that is dense in $[0,1]$. Hence, with some care taken at the boundary $0$ and $1$, it extends to a continuous strictly-increasing map from $[0,1]$ to $[A_0, A_1]$ that maps $0$ to $A_0$ and $1$ to $A_1$. That is, it extends to a homeomorphism $h$ from $[0,1]$ to $[A_0, A_1]$.
Since $h$ maps the deleted intervals during the construction of the Cantor set to the deleted intervals during the construction of $\mathscr A$, $h$ also maps the complement of the union of the former with respect to $[0,1]$ to the complement of the union of the later with respect to $[A_0, A_1]$.
That is, $h$ maps the Cantor set $\mathcal C$ to $\mathscr A=A$. 
Hence, $\mathcal C$ and $A$ are homeomorphic.
