Assume $X$ and $Y$ be nonempty subsets of $R$ such that $xWe have that both sets are nonempty, by the completeness axiom, both sup and inf exist for both sets. Since $x < y$, $y$ is an upper bound for $X$ and hence by def. of sup, sup$X \le y$. So $\sup X$ is a lower bound for $Y$ and thus by def. of inf, $\sup X \leq \inf Y$. 

Actually it's enough that, given $x \in X$ there exists an $y_x \in Y$ such that $x \leq y_x$.
In fact in this case we have $supY\geq y_x \geq x \ \forall x \in X$, which implies $supY \geq sup X$.
