I rather suspect that this choice of n was done by "trial and error"! (Many proofs are created by trial and error, then written out with showing the "errors".) Of course, it is not really the "n" that is important- it is that we want. We want to prove that if , then is not an upper bound for the set given. is the amount by which is less than 2 and the denominator was chosen so that is still less than 2.
You say "it makes sense looking at the proof" and I suspect that they just tried divided by something to get a fraction of the interval and tried different divisors until they got something that worked.
Here's how I would do that, with a slightly different "estimate".
Suppose is less than 2. Then we must have for some h. Use that h to find some number larger than whose square is still less than 2.
First note that and so we can assume that and that .
What about the number half way between? . No, that won't work- , so .
What about ? . and . Since we have to add , that still isn't good enough- we might have larger than 1 so that .
Okay, what about ? .
Since , and since , , so . Yes, we have a number, larger than whose square is still less than 2 contradicting the fact that \alpha is the least upper bound of that set.