Given that B ⊂ A. Since sup (A) exist therefore A has an upper bound sya a then a is also an upper bound of B. Therefore sup (B) exist and On the order hand for every x ⊂ A there exist y ⊂ B such that x<=y.
Therefore x<=y<=sup(B). Hence
sup(A)<=sup(B)
Therefore from (1) and (2) sup(A)=sup(B).