I'm guessing you mean #|sin(x)-sin(y)|\leq |x-y|#, since the equation you've written is not true (it's approximately true in certain situations).
Let #f(z)=sin(z)#. Let #x# and #y# be arbitrary real numbers. If #x=y#, the conclusion is obvious. Therefore, without loss of generality, we may assume that #x<y#.
Now apply the Mean Value Theorem to the function #f(z)=sin(z)# on the closed interval #[x,y]#, which can be done because #f# is definitely continuous and differentiable everywhere. From this application, we can conclude that there is a number #c\in (x,y)# such that #f'(c)=(f(y)-f(x))/(y-x)#. Since #f(x)=sin(x)#, #f(y)=sin(y)#, and #f'(c)=cos(c)#, we can rewrite this last equation as #sin(y)-sin(x)=cos(c)\cdot (y-x)#. We can now take the absolute value of both sides of this equation and use the fact that #|cos(c)|\leq 1# for any number #c# to say #|sin(y)-sin(x)|=|cos(c)||y-x|\leq |y-x|#.
Note that this is also equivalent to #|sin(x)-sin(y)|\leq |x-y|#.
