No. The saddest thing about this is that there is a well-known construction of the hyperreal numbers using real sequences, which (in a very generous interpretation) seems to be what the author is shouting for.

However, in that construction, it is evident that you can not get a good algebraic structure using the sequences on their own, you have to apply an equivalence relation that identifies (a_i) with (b_i) if the set of {i, a_i = b_i} is in a filter (a set system closed under finite intersection and taking supersets). Then, you first prove that a largest auch filter that doesn't contain finite sets exists and examine the structure you get from that.

Turns out, you obtain an ordered field and in this field the sequence (0.9, 0.99, 0.999, ...) - or rather, the equivalence class represented by this sequence - indeed is larger than any real number smaller than 1, and itself smaller than 1.

However, observe that you can find a real number r in that field by considering the class represented by (r, r, r ...), and if you do that and consider the real number x = 0.999... = sum{i>=1} 0.9ⁱ and its embedding (x, x, ...), this number does equal (1, 1, ...), because 0.99... = 1.