If you consider

then the subset

of

of elements greater than all

is non-empty (

belongs to it)

Therefore it has a minimum,

and

Indeed, if

by definition of

so

i.e.

contradiction.

Hence

and forall

by definition of

i.e.

There are various ways I guess to show that, we can proceed by induction on

:

If

nothing to show

Assume any subset of

with

elements has a max, consider

then

has a max, let's say

two cases (because the order is total):

either

then

or

and

Actually the second proof may be better than the first: it only uses the totality of the order, which is a sufficient (and necessary) condition to be able to say that any finite non-empty subset has a maximum.