We 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.
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 $supY \geq sup X$.