Exercise in Halmos set theory book.

Exercise in Halmos set theory book.

I have other question about families. At the end of the chapter of
families, there is a paragraph in the Halmos book of sets where the author
says:
Prove also with appropriate provisos about empty families that $\bigcap_i
X_i \subset X_j \subset \bigcup_i X_i$ for each index $j$ and the
intersection and the union can in fact characterized as the extreme
solution of these inclusions. This means that if $X_j \subset Y$ for each
index $j$, then $\bigcup_i X_i \subset Y$ and $\bigcup_i X_i$ is the only
set satisfying this minimality condition; the formulation for intersection
is similar.
The first part is pretty easy to prove.
Suppose that $z\in \bigcap_i X_i.$ Then for each $i\in I$, we have that
$z\in X_i$. So, clearly if $j\in I$ that means $z\in X_j$; which prove the
first inclusion, $\bigcap_i X_i \subset X_j.$
Now suppose that $z\in X_j$. Then, $z\in X_i$ for at least one $i\in I$,
i.e., $z\in \bigcup_i X_i$; which prove the second inclusion $X_j \subset
\bigcup_i X_i$
So my question in specific is about the last part, where Halmos says: "if
$X_j \subset Y$ for each index $j$, then $\bigcup_i X_i \subset Y$ and
$\bigcup_i X_i$ is the only set satisfying this minimality condition".
How do I prove the uniqueness part?
Proposition: If $X_j \subset Y$ for each index $j$. Then $\bigcup_i X_i
\subset Y$, and $\bigcup_i X_i$ is the unique set satisfying this
minimality condition
Proof:
Suppose $z \in \bigcup_i X_i$. Then there is some $i,$ such that $z\in
X_i$ but since $X_i \subset Y$ for each index, it follows that $z\in Y$.
Therefore $\bigcup_i X_i \subset Y$.
Uniqueness: ...
Does somebody have a hint? Thanks as usual.