Why does this not prove $P\neq NP$?
What Fiorini et al. show is the following:

The TSP polytope $P_n$ over $n$ points is a polytope in $\binom{n}{2}$ dimensions whose vertices correspond to all Hamiltonian cycles in $K_n$ (the complete graph on $n$ vertices). (That is, it is the convex hull of the indicator vectors of all Hamiltonian cycles.)
Suppose that $X_n$ is a polytope whose projection over the first $\binom{n}{2}$ dimensions is $P_n$, and let $d_n$ be the number of constraints needed to define $X_n$ (i.e., the number of facets of codimension 1). Then $d_n \geq f(n)$ for some function $f(n) = 2^{\Omega(\sqrt{n})}$.

In other words, they show that TSP cannot be solved using LPs in one particular way. There could be some other way of using LPs to solve TSP which isn't ruled out by their result.
For example, perhaps you could use iterative rounding to solve TSP, at each step solving an LP. This is consistent with the result of Fiorini et al.
The method in your argument is likewise not ruled out by Fiorini et al.