$$A,B\subseteq \mathbb R \quad a\in A\quad b\in B\quad \forall a,b \quad a\leq b \\ \forall \varepsilon >0 \quad b-a<\varepsilon $$ Prove there's a unqiue $c$, which is the upper bound of $A$, and the lower bound of $B$.
I've found a way to prove that $\inf(B) = \sup(A)$. But that's it, I have no idea how to prove that there's a unique $c$ which is both the upper bound of A, and that lower bound of B.
On
It seems like your job is to prove that there exists a unique real number $C$ such that $C$ is an upper bound for $A$ and $C$ is a lower bound for $B$.
This breaks into two steps: proof of existence; and proof of uniqueness.
To prove existence, you write down a formula $C =$ SOMETHING, and you verify that this $C$ that you wrote down is an upper bound for $A$ and is also an upper bound for $B$. Your question contains the required formula: $C = \sup(A)=\inf(B)$; it is an upper bound for $A$ becaue $\sup(A)$ is an upper bound for $A$; and it is a lower bound for $B$ because $\inf(B)$ is a lower bound for $B$.
To prove uniqueness, you assume that you have another number, say $C'$, and you also assume that $C'$ is an upper bound for $A$ and is an upper bound for $B$; then, using these assumptions, you prove that $C'=C$. This is a proof by contradiction: assuming that $C' \ne C$, it follows that either $C' < C$ or $C' > C$, and so you break into a proof by cases. If $C' < C$ then, using $C = \sup(A)$, it follows that $C'$ is an upper bound for $A$ which is less than the least upper bound, contradiction. If $C' > C$ then, similarly using $C = \inf(B)$, it follows that $C'$ is a lower bound for $B$ which is greater than the greatest lower bound, contradiction.
I can't comment but here is a hint: use the property of inf and sup. You have shown $C =\inf B=\sup A$. Will anything less than $C$ be a lower bound?