What you are describing isn't even a proper binary search, it just gives you a lower bound that is guaranteed to be no more than 50% smaller than the solution.

Let's assume doing an actual binary search. What you are getting, in every step of binary search using your oracle, is an upper bound U (D(TSP, U) ==1), for which we know that there is a solution smaller or equal that U, and a lower bound L (D(TSP, L) == 0) for which we know there isn't. This is where your confusion between finite and integer-valued comes into play: If the problem were integer valued, you would have the optimal solution S as soon as L_i=U_i (implying U_i is S), which would be guaranteed after a polynomial number of steps. With real valued weights, all you get is a sequence of intervals (L_1, U_1], (L_2, U_2], ... where the size of the intervals is halved in every step. What you actually want is the exact value of the solution. For that, you would need to show that there is no hamiltonian path between L_i and U_i, which isn't something your oracle can determine.

Note that this probably doesn't show that a polynomial oracle for the TSP decision problem will not yield a polynomial solution to the optimization problem in general, it just deals with your proposition. Edit: Not -> Note