This is from Fitzpatrick's Advanced Calculus. I am not very experienced with proofs so could someone tell me if this is a correct way of doing it? Thanks!

If is a set of real numbers that is bounded above and is a nonempty subset of , then .

Claim:

Proof:Let be a set of real numbers that is bounded above and a nonempty subset of . Take an arbitrary element in . Since , is also in . Since is bounded above, by the completeness axiom we know there is a number such that , and there is no number less than that is an upper bound for . Since is in , is also an upper bound for . So we must have , or .