A cardinal number $\alpha$ is a ordinal number such that $\forall \beta < \alpha : \beta \not\approx \alpha$. Now I want to show that for a cardinal number $\kappa$ the set $\kappa \times 2$ with the lexicographic order $<_{lex}$ defined by $(a,b) <_{lex} (a',b') :\iff a < a' \lor (a' = a \land b < b')$ is order isomorph to $\kappa$ with the well order $\in$.
So far I was able to show that $\forall \lambda < \kappa : \kappa \times 2 \approx \lambda$ and for infinite ordinal numbers $\alpha$ one has $\kappa \times 2 \approx \alpha$.
Edit: By transfinite induction on $\kappa$. We know $\omega \times 2 \approx \omega$ and $|\omega \times 2| = \omega$. Thus we can assume that it holds for smaller infinite cardinals $\lambda < \kappa$ und prove it for $\kappa$. My Idea is to note that $<_{lex}$ is a well ordering on $\kappa \times 2$ and starts out by \begin{equation} (0,0) <_{lex} (1,0) <_{lex} (2,0) <_{lex} (3,0) <_{lex} \ldots <_{lex} (\omega,0) <_{lex} (\omega,1). \end{equation} Now each $(a,b) \in \kappa \times 2$ has $|\max(a,b) + 1 \times \max(a,b) + 1| < \kappa$ many $<_{lex}$ predecessors and $otp((\kappa \times 2)) = \kappa$ hence there must be a order isomorphism between $(\kappa \times 2, <_{lex})$ and $(\kappa, \in)$.
What you want to show is that if $\kappa$ is a cardinal then it is the unique order type of a well-order that has cardinality $\kappa$ and every proper initial segment has size $<\kappa$.
So, for example, $\omega$ is the unique well-ordering that is countable and has every proper initial segment finite. And $\omega_1$ is the unique order type of a well-order which is uncountable and every proper initial segment is countable.
So, to prove that $\kappa\times2$ is isomorphic to $\kappa$, it is enough to show that any proper initial segment has size $<\kappa$, as very easily $\kappa\times2$ has cardinality $\kappa$.
For this, show that if $A\subseteq\kappa\times2$ is a proper initial segment, then there is some $\gamma<\kappa$ such that $A\subseteq\gamma\times2$.