See (my) writeup in monotone sequence theorem, or none of this will make sense. Also, be warned! SPOILERS AHEAD!

Well, suppose our sequence a_{1},a_{2},... has *no* infinite ascending subsequence. Let's use this to find an infinite descending subsequence!

Consider a "chain" ascending from a_{1} (l_{1,1}=1): a_{l1,1}<a_{l1,2}<a_{l1,3}<...
(we'll see why we have the first index in a moment...). Since there's no infinite ascending subsequence, this chain must end at some last element a_{k1}, so we know that a_{k1}≥a_{j} for any j>k_{1}.

OK, so now take l_{2,1}=k_{1}+1, and extend an ascending chain from a_{l2,1} as far as possible (again, by assumption it's finite!). Call the last element this time k_{2}. Then k_{2}>k_{1}, so we have a_{k1}≥a_{k2}. On the other hand, a_{k2}≥a_{j} for any j>k_{2}.

Continue in this fashion: always pick l_{j,1}=k_{j-1}+1, then a maximal ascending sequence a_{lj,1}<a_{lj,2},..., conclude that it must be finite, and take a_{kj} as its last element.

This gives us an infinite descending subsequence, a_{k1}≥a_{k2}≥..., as required for a proof of the theorem.