I'm not sure that I follow your induction step. It looks way too convoluted, though. There are two proofs of this I like. The first is more direct, and the second invokes a topology flavor which is really cool.
A) The point to notice is that if x ∈ A and y ∈ A, with x > y, then x - y ∈ A. So, if the infimum is a positive number c, there are two possibilities. Either c is in A or c is not in A. If c is not in A, then there is an infinite number of elements of A, between c and 2c. Subtracting two of them we get a value smaller than c in A, contradiction.
Then, c is in A. But that means that the minimum distance between two elements of A is c, and that all numbers of the form nc, for n ∈ ℕ, are also in A. It's straightforward to see that the elements of the form nc are the only ones allowed, because if there were another one called x, sitting between nc and (n+1)c, then either x-nc or (n+1)c-x, would give a number smaller than c, contradicting the fact that c is the minimum.
Therefore, if the infinimum is positive, all elements of A are of the form nc, but that means that w is rational. So, inf A = 0
B) To prove it topologically, the key point to notice is that, for an irrational number w, the fractional parts of any multiple nw should be all different. If two of them were the same, we would have nw - mw = k, an integer, and w = k/(n-m) a rational number, which is absurd.
Therefore, the sequence nw-floor(nw) is a subset of A, has an infinite number of distinct elements, and is also bounded, since all elements are between 0 and 1. Therefore, we can apply the
https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem and find a convergent subsequence. This means that, for any epsilon, we can find two elements of A x and y such that x - y < epsilon. Since x - y ∈ A, we just proved that there are elements of A which are arbitrarily small, which is enough to prove that inf A = 0.