${\rm sup}\ A\cap B = {\rm min}\ \{ {\rm sup} (A), {\rm sup}(B) \} $ 
Let $A,B\subseteq \mathbb{R}$ be a non-empty intervals and bounded from above.
  
  If $A\cap B\neq \emptyset $ prove that it is bounded from above and that $Sup(A\cap B)=min\{sup(A),Sup(B)\}$

$A,B$ are bounded from above therefore there are $M_1,M_2$ such that $a\leq M_1$ and $b\leq M_2$.
 for $x \in A\cap B$ $(x\in A \wedge x\in B)$, so  $x\leq M_1 \wedge x\leq M_2$. 

W.L.O.G let say that $A\leq B$ if so than $x\leq M_1\leq M_2$, but there is only one Supremum and by definition it is the least upper bound therefore   $Sup( A\cap B)= M_1$ and if $B\leq A$ the proof is the same.

Is it a valid proof?
 A: The statement is not true. Think of 
$$A = \{1, 2\}, B = \{1, 3\}. $$
A: Since $A=(a_0, a_1)$ and $B=(b_0, b_1)$ with $a_1, b_1 < \infty$ and $a_0, b_0 \ge -\infty$, write out the intersection explicitly:
$$A\cap B = (\max(a_0, b_0), \min(a_1, b_1))$$
The supremum of an interval is its right endpoint so
$$\sup A\cap B = \min(a_1, b_1) = \min(\sup A, \sup B)$$
Note that it's irrelevant for this proof if the intervals have open or closed ends at any side.
A: In general, we can show that the sup of a nonempty intersection is smaller than or equal to the sup of either of the sets:  $$ (1)-sup(A\cap B)≤sup(A)$$ or $$(2)-sup(A\cap B)≤sup(B).$$ Proof of (1): $ $ Let x $\in$ (A $\cap$ B) be arbitrary. Then x ≤ sup(A $\cap$ B). Since x $\in$ (A $\cap$ B), x $\in$ A, so x ≤ sup(A). Thus, sup(A) is an upper bound for A $\cap$ B. However, since A $\cap$ B $\subset$ A and sup(A $\cap$ B) is the least upper bound for A $\cap$ B, it follows that sup(A $\cap$ B) ≤ sup(A).  Proof of (2): $ $ Let y $\in$ (A $\cap$ B) be arbitrary. Then y ≤ sup(A $\cap$ B). Since y $\in$ (A $\cap$ B), y $\in$ B, so y ≤ sup(B). Thus, sup(B) is an upper bound for A $\cap$ B. However, since since A $\cap$ B $\subset$ B and sup(A $\cap$ B) is the least upper bound for A $\cap$ B, it follows that sup(A $\cap$ B) ≤ sup(B). Of course, in both proofs we are assuming both A and B are subsets of $\mathbb{R}$ and that both A and B are bounded above in order for sup(A) and sup(B) to exist.
