I do not know the best method of doing this. This is what I have done:
(x.y)' = x' + y'
↔
(x.y)' + x.y = x' + y' + x.y ............ (assuming x.y != 1)
↔
1 = x' + y' + x.y
↔
1 = x' + (y' + x).(y' + y)............... (Distributive property)
↔
1 = x' + (y' + x)
↔
1 = 1
Now, in the first step we assumed that x.y != 1. If it was so, then the statement is obviously true.
P.S.: I am myself not fully satisfied with this proof as we still deal with it in cases. It's not one-blow-for-all!