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Lets Assume to the contrary that there is a Supremum $S_1 \in A \cap B$ that is larger than the Supremum $S_2 \in A$

If $S_1 \in A \cap B$ then $S_1 \in A$ and since $S_2 \in A$

This is a contradiction because $S_1 \not\gt S_2$

and therefore sup$(A \cap B) \leq$ sup$(A)$

Is this sufficient to disprove this? I could use a counter example but since its contradictory to have a smallest upper bound in a set larger than the smallest upper bound of a set I wanted to go this way.

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    What if $S_1\notin A\cap B$? – 2017-02-01
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    A supremum (or infimum also) of a set need not be in the set. – 2017-02-01
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    Note that if $\sup A \leq \sup (A\cap B)$, then $\forall a\in A \ \exists b\in A\cap B$ that satisfies $b\geq a$. – 2017-02-01

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Don't make things unnecessarily complicated. You don't need a proof by contradiction here.

By definition $S=\sup A$ is an upper bound for all elements of $A$, hence an upper bound for all elements of $A\cap B$. By definition $\sup(A\cap B)$ is the smallest of all upper bounds for the elements of $A\cap B$. Thus $S\ge \sup(A\cap B)$.