Show that for all x and y, absolute value of sin(x)-sin(y) is less than or equal to the absolute value of x-y
\[\left| \sin(x)-\sin(y) \right|\le \left| x-y \right|\]

 Ok, f(z) is everywhere continuous and differentiable, hence MVT is applicable. 

MVT says for every x,y, 

f'(z) = cos(z) = (sin(x)-sin(y)) / (x-y) 

But you know -1 ≤ cos(z) ≤ 1. i.e. 0 ≤ |cos(z)| ≤ 1 

=> -1 ≤ (sin(x)-sin(y)) / (x-y) ≤ 1 

Multiply across by (x-y), and note the inequality remains valid for both the +ve and -ve cases due to the symmetry (i.e. -1 swaps with 1 and v.v.): 
=> -(x-y) ≤ (sin(x)-sin(y)) ≤ (x-y) 

Take || (abs) of both sides. 

=> 0 ≤ | sin(x)-sin(y) | ≤ | x-y | 

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The correct answer would be "convenience"