Proof of Bedrosian's theorem
One easy way to see the validity of the theorem is to consider the corresponding analytic signals. Let a signal $s(t)$ be defined by
$$s(t)=m(t)c(t)\tag{1}$$
where $m(t)$ is a low pass (message) signal, and $c(t)$ is a bandpass (carrier) signal. Note that we don't need to assume that $c(t)$ is sinusoidal. Let $\hat{x}(t)$ denote the Hilbert transform of $x(t)$, and $\hat{c}(t)$ is the Hilbert transform of $c(t)$.
The equality claimed by Bedrosian's theorem
$$\hat{s}(t)=m(t)\hat{c}(t)\tag{2}$$
is equivalent to
$$s_a(t)=m(t)c_a(t)\tag{3}$$
where $s_a(t)$ and $c_a(t)$ denote the analytic signals corresponding to $s(t)$ and $c(t)$, respectively. Analytic signals are complex-valued signals having only positive frequency components. The analytic signal is given by the original signal plus its Hilbert transform as the imaginary part:
$$s_a(t)=s(t)+j\hat{s}(t)\tag{4}$$
In the frequency domain, the signal $m(t)c_a(t)$ is centered at the (positive) center frequency of the bandpass signal $c(t)$. Its bandwidth equals the bandwidth of $c(t)$ plus twice the bandwidth of the lowpass signal $m(t)$. It has only positive frequency components if the spectra of $m(t)$ and $c(t)$ do not overlap. In that case it equals the analytic signal $s_a(t)$, as claimed in Eq. $(3)$. Otherwise, $m(t)c_a(t)$ has also negative frequency components and $(3)$ is not satisfied.
Note that $(2)$ and $(3)$ are also valid if $c(t)$ is a highpass signal, as long as the spectra of $m(t)$ and $c(t)$ don't overlap.
Also take a look at this related answer and the figure therein.