Applying duality property to fourier transform of unit step function
The result you got is correct and it is also expected according to the two first formulas in your question. If $X(f)$ is the Fourier transform of $x(t)$, then the Fourier transform of $X(t)$ equals $x(-f)$. If $x(t)=u(t)$ is the unit step function then

$$X(f)=\frac12 \delta(f)+\frac{1}{j2\pi f}\tag{1}$$

and the Fourier transform of $X(t)$ is given by

$$\mathcal{F}\left\{\frac12 \delta(t)+\frac{1}{j2\pi t}\right\}=\frac12+\frac{1}{2 j}\mathcal{F}\left\{\frac{1}{\pi t}\right\}\tag{2}$$

where we recognize $1/\pi t$ as the impulse response of an ideal Hilbert transformer, the Fourier transform of which is given by

$$\mathcal{F}\left\{\frac{1}{\pi t}\right\}=-j\;\textrm{sgn}(f)\tag{3}$$

Combining $(2)$ and $(3)$ gives

$$\mathcal{F}\left\{X(t)\right\}=\frac12\big(1-\textrm{sgn}(f)\big)=u(-f)\tag{4}$$

just as expected.

Maybe you can clarify which books say otherwise.