Though the way you asked for it is confused, I believe the argument you need is this one:
The key result is:
If $A$ is a set and $\scr A$ is a collection of subsets of $A$ such that there is a one-to-one correspondence between $A$ and $\scr A$, then there is some subset $B \subseteq A$ with $B\notin \scr A$.
If $a \in A$, then let $S(a)$ be the corresponding element of $\scr A$. Define $B$ by $$B = \{a \in A\mid a \notin S(a)\}$$
Suppose $C \in \scr A$. There must be some $c \in A$ with $S(c) = C$. There are two possibilities:
- if $c \in C$, then $c \in S(c)$ and therefore $c \notin B$ by definition. Therefore $C \ne B$ since $C$ contains an element that $B$ does not contain.
- if $c \notin C$, then $c \notin S(c)$ and therefore $c \in B$ by definition. Once again, $C \ne B$, since $B$ contains an element that $C$ does not contain.
Thus $B$ is not equal to any element of $\scr A$. I.e., $B \notin \scr A$. QED.
This is actually a generalization of Cantor's diagonal argument. $\big(c,S(c)\big)$ represents the diagonal of $A \times \scr A$, and we've chosen $B$ to disagree with each entry along the diagonal. You may also note I made no use of the one-to-one-ness of the correspondence. It actually works for any correspondence between $A$ and $\scr A$ where each element of $A$ corresponds to only one set in $\scr A$, and every element of $\scr A$ corresponds to at least one element of $A$ (a "surjection from $A$ to $\scr A$").
The set $\mathscr P(A) = \{C \mid C \subseteq A\}$ of all subsets of $A$ is called the "power set of $A$". A corollary of the above result is
There can be no one-to-one correspondence between $A$ and $\mathscr P(A)$.
If there were, there would be some $B \subseteq A$ with $B \notin \mathscr P(A)$. But $\mathscr P(A)$ contains every subset of $A$, so that cannot be. But $A$ is obviously in one-to-one correspondence with the set $\big\{ \{a\}\mid a \in A\big\} \subset \mathscr P(A)$. Thus we can say
For any set $A$, $|A| < |\mathscr P(A)|$
where $|A|$ indicates the cardinality of $A$. But $\mathscr P(A)$ is also a set, so we can say $$|\mathscr P(A)| < |\mathscr P(\mathscr P(A))|.$$
In particular,
$$|\Bbb N| < |\mathscr P(\Bbb N)| < |\mathscr P(\mathscr P(\Bbb N))| < |\mathscr P(\mathscr P(\mathscr P(\Bbb N)))| < \dots$$
Thus there must be infinitely many infinite cardinals.
Now $\aleph_0 = |\Bbb N|$ by definition. But also by definition $\aleph_1$ is the smallest cardinal $> \aleph_0$ (any non-empty collection of cardinals will always have a smallest element). Similarly $\aleph_2$ is defined to be the smallest cardinal $> \aleph_1$, and so on.
So as Asaf Karagila pointed out, the strict answer to your question is $\aleph_2$ cannot be put into one-to-one correspondence with $\aleph_1$ because by definition it is a larger cardinal that $\aleph_1$. What the argument above shows is that there actually is a cardinal $\aleph_1 > \aleph_0$ and actually is a cardinal $\aleph_2 > \aleph_1$, etc.
We know that $|\Bbb R| = |\mathscr P(\Bbb N)|$. What we do not know (and what makes your question confused) is whether $\aleph_1 = |\mathscr P(\Bbb N)|$. It must be that $\aleph_1 \le |\mathscr P(\Bbb N)|$, since it is the smallest cardinal above $\aleph_0 = |\Bbb N|$. But whether or not there are any cardinals between $\aleph_0$ and $|\mathscr P(\Bbb N)| = |\Bbb R|$ is actually an undecidable question. You can assume it to be true or assume it to be false, and never run into a contradiction either way.
