MHB Minimal successor set - difficult

[Andrei1]

Prove that for all $$x,y\in\omega,\ \ x\subset y\vee y\subset x.$$

If I assume that the conclusion is false then I can prove that for some $$a\in x,\ b\in y$$ we have $$a\notin b$$ and $$b\notin a.$$

Also I am thinking that if assume the contrary then $$\omega$$ minus $$\{x\}$$ or minus $$\{y\}$$ or both is a smaller successor set. Should I try to prove this?

I get stuck in trying to prove for sets from $$\omega$$ the equivalence: $$a\subseteq b\wedge a\not=b\Leftrightarrow\exists c(a\cup c^+=b)$$.

 

[Andrei1]

Here's the plan.
(1) Prove $$a\subset b^+\Rightarrow b\notin a$$ by induction on $$a.$$ Use also $$x=y\Rightarrow x^+=y^+.$$
(2) Prove $$a\subseteq b\Leftrightarrow a\subset b^+$$. In proving ($$\Leftarrow$$) side use (1). In proving ($$\Rightarrow$$) side use $$x\subset x^+$$, which follows from $$x\notin x.$$
(3) Prove $$b\subset a\Leftrightarrow b^+\subseteq a$$. In proving ($$\Rightarrow$$) side use induction on $$a.$$ Use also $$x=y\Rightarrow x^+=y^+$$ and $$x\subseteq x^+.$$ In ($$\Leftarrow$$) side you need $$x\notin x.$$
(4) Prove $$a\subset b\vee a=b\vee b\subset a$$ by using (2) and (3). Use induction.