If $X \setminus Y$ is empty, then the claim holds due to R7: Empty set is subset of every set.

Suppose thus that $X \setminus Y$ is not empty and let $x \in X \setminus Y$. Then $x \in X$ and $x \not\in Y$. Since $Z$ is a subset of $Y$, it must be true that $x$ is not in $Z$ either. But now, by definition, $x \in X \setminus Z$. Since $x \in X \setminus Y$ was arbitrary, the claim follows. $\square$