For steps 1 and 2, perhaps you have the right intuition (I don't know), but anyway I really don't understand what you wrote. Here are a few hints to let you improve your proof (I give you the idea, I let you formalize the proofs):

A floor term is such that: , . In words, it is less than or equal to any of the following terms in the sequence.

1st case: there are infinitely many floor terms. Each floor term is less than or equal to every following term, and in particular it is less than (or equal to) the following floor term. So that the floor terms form an increasing subsequence.

[To write the proof, I suggest you begin by naming the indices of the floor terms, for instance , : is the first floor term, for the second one, and so on]

2st case: there are finitely many floor terms (and at least one). Let be the last one. Then, after , there is no floor term.

What does it mean not to be a floor term? It means that you can find a (strictly) smaller term among the following ones.

So there is a term strictly smaller than (after ), and again there is a term strictly smaller than this one, and so on... This constructs a (strictly) decreasing subsequence.

[This is a construction by recursion: the first term is . Suppose you defined the first terms of the subsequence, say where , then define which term is ]

3rd case: there is no floor term. This is no different from the 2nd case (it could be possible to gather 2nd and 3rd cases). Every term is a non-floor term, which means that there's a stricly smaller term later in the sequence. Apply this to , then to the term smaller than , and so on and so forth.

I hope it clarifies the way the proof works.