How to prove: If $A_2 ⊆ A_1 ⊆ A_0 $ and $A_2 ≈ A_0$ then $A_2 ≈ A_1 ≈ A_0$ $A_0$ is a set, "$A_0 ≈ A_2$" means there is a bijective function from $A_0$ to $A_2$. So, how to prove the following proposition?
If $A_2 ⊆ A_1 ⊆ A_0$ and $A_2 ≈ A_0$ then $A_2 ≈ A_1 ≈ A_0$.
 A: As Jens Schwaiger noted in the comments, this follows easily from the Cantor-Schröder-Bernstein theorem, and I’ve not found a direct proof that isn’t in effect reproving that theorem. This is hardly surprising, since the Cantor-Schröder-Bernstein theorem follows easily from the result in the question. But for what it’s worth, here is a sketch of such a proof.
Let $f:A_0\to A_2$ be a bijection. Let $D_0=A_0\setminus A_1$ and $D_1=A_1\setminus A_2$. For $n\ge 1$ let $A_{n+2}=f[A_n]$; an easy induction shows that $A_n\supseteq A_{n+1}$ for all $n\ge 0$. For $n\ge 0$ let $D_{n+2}=f[D_n]$; $$D_2=f[D_0]=f[A_0\setminus A_1]=f[A_0]\setminus f[A_1]=A_2\setminus A_3\;,$$ and another easy induction establishes that in general we have $D_n=A_n\setminus A_{n+1}$ for $n\ge 0$. Moreover, the family $\{D_n:n\ge 0\}$ is easily seen to be pairwise disjoint.
Let $$A_\infty=\bigcap_{n\ge 0}A_n\;.$$
Then
$$A_0=A_\infty\sqcup\bigsqcup_{n\ge 0}D_n=A_\infty\sqcup\bigsqcup_{n\ge 0}D_{2n}\sqcup\bigsqcup_{n\ge 0}D_{2n+1}\;,$$
and
$$A_1=A_\infty\sqcup\bigsqcup_{n\ge 1}D_n=A_\infty\sqcup\bigsqcup_{n\ge 1}D_{2n}\sqcup\bigsqcup_{n\ge 0}D_{2n+1}\;,$$
where $\sqcup$ and $\bigsqcup$ denote disjoint union. Define
$$g:A_0\to A_1:a\mapsto\begin{cases}
a,&\text{if }a\in A_\infty\sqcup\bigsqcup_{n\ge 0}D_{2n+1}\\
f(a),&\text{if }a\in\bigsqcup_{n\ge 0}D_{2n}\;;
\end{cases}$$
then $g$ is a bijection from $A_0$ onto $A_1$, so $A_0\approx A_1$. Finally, $g\circ f^{-1}$ is a bijection from $A_2$ to $A_1$, so $A_2\approx A_1$.
