Where is the flaw in this derivation of the DTFT of the unit step sequence $u[n]$?
There are infinitely many signals that make the following equality hold: $$y[n]-y[n-1]=\delta[n] \qquad (1)$$
The only thing that matters is that $y[0]-y[-1]=1$, and then the rest of the coefficients of $y$ can be determined under the restriction that Eq. $(1)$ states (i.e. the substraction of consecutive samples must be $0$ for $n\neq 0$).
In other words, Eq. $(1)$ will be achieved by any signal $y[n]$ such that $$y[0]=y[-1]+1 \land y[n]=y[n-1] \ \forall n\neq0$$
Another way to see this is that any function that is basically $u[n]$ with an offset (a constant value added) will satisfy $(1)$. This explains the statement made by robert bristow-johnson in his answer: differentiators destroy this information (such as taking a derivative in continuous time destroys evidence of any constant value in the original function).

To sum up, I believe that the proof is flawed because the procedure followed could use any function of the form $u[n]+C$ with $C\in\mathbb{R}$, and this would lead to many functions having the same Fourier transform, which is indeed wrong as the Fourier transform is a bijection. Maybe the author deliberately decided to ignore anything related to DC values, conscious that in order to show that $F(\omega)$ is the DTFT of $f[n]$ he would need the accumulation property (whose most popular proof is derived from the DTFT of the unit step - ergo, a pretty circular proof). The proof is not strictly wrong, as everything it states (the formulae for $F(\omega)$ and $U(\omega)$, the decomposition of the unit step, the difference equation) is true, but it would require the accumulation property to show why $F(\omega)$ doesn't have any Dirac deltas.