Prove Trig Identity 
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$

This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\right)}^y2\cos\left(\frac{B-C}2\right)\\
\sin\left(\frac{A-C}2+\frac{A-B}2\right)+\sin\left(\frac{A-C}2-\frac{A-B}2\right)\\
\left(\frac{B-C}2\right)\left[\sin\left(\frac{2A-B-C}2\right)+\sin\left(\frac{B-C}2\right)\right]\\
\cos\left(\frac{B-C}2\right)\sin\left(\frac{2A-B-C}2\right)+\cos\left(\frac{B-C}2\right)\sin\left(\frac{B-C}2\right)$$
I know it's basically turning sum into product and I previously tried to derive the LHS the same way the normal sum to product identity works but I couldn't figure out how. I tried to use the sum-product identity on RHS but couldn't quite make it work.

You are missing a factor of $2$ in the last line of your answer. Anyway, you were on the right track. You just need to use the same type of identity (as you used in the beginning) for the first term and the double angle formula for the second.
$$
2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{2A - B - C}{2}\right) + 2\cos\left(\frac{B-C}{2}\right)\sin\left(\frac{B-C}{2}\right) \\
= \sin\left(\frac{B-C}{2} + \frac{2A - B - C}{2}\right) - \sin\left(\frac{B-C}{2} - \frac{2A - B - C}{2}\right) + \sin(B - C) \\
= \sin(A - C) - \sin(-A + B) + \sin(B - C) \\
= \sin(A - C) + \sin(A - B) + \sin(B - C)
$$
