Is This Proof for "If $\sup A < \sup B$, show that there exists an element $b\in B$ that is an upper bound for $A$" correct? This question is from Understanding Analysis (Stephan Abbot) Exercise $1.3.9$. The Question is  If $\sup A < \sup B$, show that there exists an element $b \in B$ that is an upper bound for $A$. My proof is as follows:
If there exists an element $b\in B$ that is an upper bound for $A$ than $(\exists b\in B)(\forall a\in A) a <b$. Assume (for the sake of contradiction) that $\sup(A) < \sup(B)$ but $(\forall b\in B)(\exists a\in A)b \leq a$. Since $\sup(A) \geq a \geq b (\forall a \in A, b \in B)$ and $\sup(A) < \sup(B)$, $\sup(A)$ is an upper bound for B which is less than $\sup(B)$ which is a contradiction. Therefore, if $\sup A < \sup B$, there exists an element $b\in B$ that is an upper bound for $A$.
Is this proof correct?

Your proof is correct but be careful with quantifiers.  $(\forall a \in A ,b \in B\, a \geq b)$ is not true. We only know that for each $b \in B$ there exists $a \in A$ such that $a \geq b$, but in general this is not true for every element $a \in A$.
Here is another proof which I find be more natural:
Let $c \in ]\sup A, \sup B[$. There exists $b \in B$ such that $c \leq b$, and thus for all $a \in A$ we have
$a \leq \sup A \leq c \leq b$
so $b$ is an upper bound for $A$.
You will find that this proof is rather intuitive when making a drawing of the situation.
