Suppose that $X$ is contained in $Y$. If $X \setminus Z$ is empty, then R7: Empty set is subset of every set guarantees that $X \setminus Z$ is a subset of $Y \setminus Z$.

Suppose then that $X \setminus Z$ is not empty and let $x \in X \setminus Z$. Now, first and foremost, $x$ belongs to $X$. Since $X$ is contained in $Y$, then $x$ belongs also to $Y$. Because we also assumed that $x \not\in Z$, it follows that $x \in Y \setminus Z$. Since $x \in X \setminus Z$ was arbitrary, the claim follows. $\square$