Proof checking: $B\subseteq A$ implies $\sup B \leq \sup A$

Question

Given $B \subseteq A$. Assume $\sup B > \sup A$. Now $\forall \epsilon >0$ $\exists b\ \in B (b > \sup B - \epsilon)$. Take $\epsilon = \sup B- \sup A$. That is $\exists b\ \in B (b > \sup A)$.

Since $b$ also belongs to $A$. So we have found some element of $A$ which is bigger than its Supremum. Hence a contradiction.

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