Proof of fourier transformation of multiplication of two signals
I am very thankful to Dilip-sarwate and Gilles, who took their precious time to understand my problem and guide me.
So, Now I'm going to write the correct solution to my question. Which is as follows:
 $$
ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}[x(t)g(t)]e^{-j\omega t}dt
$$
As we know : 
$$
x(t) = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha 
$$
Therefore,
$$
ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}\bigg[\frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha\bigg]g(t) e^{-j\omega t}dt
$$
After rearranging, we have
$$
ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)\bigg[\int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt\bigg]d\alpha
$$
As we also know that :
$$
G(\omega) = \int_{t=-\infty}^{\infty}g(t)e^{-j\omega t}dt
$$
and replacing $\omega$ with $\omega-\alpha$, we get:
$$
G(\omega-\alpha) = \int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt
$$
Therefore,
$$ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)G(\omega-\alpha)d\alpha$$
Finally, we get
$$
ℱ[x(t)g(t)] = \frac 1{2\pi}[X(\omega)*G(\omega)]
$$

[$\because$ $x_1(t)*x_2(t) = \int_{\tau=-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau$]