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, $\sup X \leq \inf Y$. 

That is basically a good proof, but could use some clarifications.  In particular you should try to distinguish more clearly between numbers and sets.


*

*sup and inf do not necessarily exist for a nonempty set of real numbers.  However, it is true that a nonempty set which has an upper bound has a sup, and similarly a nonempty set which has a lower bound has an inf, and that's clearly true here.

*When you say $x < y$, you're considering a single $y$ and all $x \in X$ to conclude that sup $X$ $\le y$.

*Because sup $X$ $\le y$ is true for all $y \in Y$, we can also conclude that sup $X$ is a lower bound for $Y$.  It is not a lower bound for $y$: we say that sets have bounds, but numbers do not.
