ultrafilters as linear orders

The first part is just an easy application of Łoś's theorem but is trivially proved directly.

The second part. If $a=((n))$ is an element of the diagonal image of $A$ and $b=((b_i))<a$, then almost for all $i$ $b_i<n$. Then there are only finitely many different $b_i$. Since the ultrafilter is not principal, there is a set $S$ from the ultrafilter and $s<n$ such that $b_i=s$ for all $i\in S$. Then $b=((s))$ belongs to the image of the diagonal embedding. QED

The third part is straightforward: just find an infinite descendent sequence starting with $(1,2,2^2, 2^{2^2}, 2^{2^{2^2}},...)$ (again using the fact that the ultrafilter is not principal).