general topology – Union of open sets is open

The proof is generally pretty straightforward, but my only difficulty is with a small subtlety. The statement is:

Let $(X,d)$ be a matrix space and ${U_i}_{i in I}$ a collection of open sets. Prove that $bigcuplimits_{i in I} U_i$ is open.

My attempt is:

If $I = emptyset$, then $bigcuplimits_{i in I} U_i = emptyset$, which is open, so we proceed under the assumption that $I neq emptyset$. Furthermore, if $U_i = emptyset$ for all $i in I$, then $bigcuplimits_{i in I} U_i = emptyset$, which is open, so assume $U_i neq emptyset$ for at least one $i in I$. So, let $x in bigcuplimits_{i in I} U_i$. Then $x in U_{i_0}$ for some $i_0 in I$. As $U_{i_0}$ is open, there exists $epsilon > 0$ such that $B_{epsilon} (x) subset U_{i_0} subset bigcuplimits_{i in I} U_i$, so $x$ an interior point of $bigcuplimits_{i in I} U_i$. Therefore, $bigcuplimits_{i in I} U_i$ is open.

Every proof of this fact I have every seen completely omits the first two cases I wrote down, where $I = emptyset$ or $U_i = emptyset$ for all $i in I$. Are these even necessary to consider? Are they just too trivial to comment on?