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 $\displaystyle S$ is a set of real numbers that is bounded above and $\displaystyle B$ is a nonempty subset of $\displaystyle S$, then $\displaystyle \text{sup }B\leq \text{sup }S$.

Claim:

Proof:Let $\displaystyle S$ be a set of real numbers that is bounded above and $\displaystyle B$ a nonempty subset of $\displaystyle S$. Take an arbitrary element $\displaystyle x$ in $\displaystyle B$. Since $\displaystyle B\subseteq S$, $\displaystyle x$ is also in $\displaystyle S$. Since $\displaystyle S$ is bounded above, by the completeness axiom we know there is a number $\displaystyle c=\text{sup }S$ such that $\displaystyle x\leq c$, and there is no number less than $\displaystyle c$ that is an upper bound for $\displaystyle S$. Since $\displaystyle x$ is in $\displaystyle B$, $\displaystyle c$ is also an upper bound for $\displaystyle B$. So we must have $\displaystyle \text{sup }B\leq c$, or $\displaystyle \text{sup }B\leq \text{sup }S$.