Hilbert transform pair proof
I agree that one of the easiest ways to compute the Hilbert transform in this case is to use the analytic signal. This is most easily obtained via the Fourier transform. Note that the Fourier transform of

$$f(t)=\frac{\sin(at)}{at}\tag{1}$$

is

$$F(j\omega)=\frac{\pi}{a}\text{rect}\left(\frac{\omega}{2a}\right)\tag{2}$$

i.e., a rectangular function with value $\pi/a$ in the interval $[-a,a]$, and zero otherwise. The spectrum of the corresponding analytic signal equals the part of $F(j\omega)$ at positive frequencies, scaled by a factor of $2$:

$$F_a(j\omega)=\frac{2\pi}{a}\text{rect}\left(\frac{\omega}{a}-\frac12\right)\tag{3}$$

The inverse Fourier transform of $(3)$ gives the analytic signal:

$$f_a(t)=\frac{1}{2\pi}\frac{2\pi}{a}\int_0^{a}e^{j\omega t}d\omega=\frac{1}{jat}\left(e^{jat}-1\right)=e^{jat/2}\frac{\sin(at/2)}{at/2}\tag{4}$$

The Hilbert transform of $f(t)$ is the imaginary part of $(4)$:

$$\hat{f}(t)=\text{Im}\{f_a(t)\}=\frac{\sin^2(at/2)}{at/2}\tag{5}$$

as stated in your question.