Result
R4145: Binary union is an upper bound to both sets in the union shows that $X, Y \subseteq X \cup Y$. Result
R4159: Power set is not empty shows that $\mathcal{P}(X) \cup \mathcal{P}(Y)$ is not empty, so we need not consider the case where $\mathcal{P}(X) \cup \mathcal{P}(Y)$ is empty. If $E \in \mathcal{P}(X) \cup \mathcal{P}(Y)$, then $E \subseteq X$ or $E \subseteq Y$. Since the labels are arbitrary, we may assume without loss of generality that $E \subseteq X$. Then $E \subseteq X \subseteq X \cup Y$, so that $E \in \mathcal{P}(X \cup Y)$. Since $E \in \mathcal{P}(X \cup Y)$ was arbitrary, the proof is complete. $\square$
