Component of Finite Union in Ultrafilter

Theorem
Let $S$ be a non-empty set.

Let $\mathcal U$ be a ultrafilter on $S$.

Let $\left\{{Y_1, \dots, Y_n}\right\}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$.

Then there is a unique $k \in \left\{{1, \dots, n}\right\}$ such that $Y_k \in \mathcal U$.

Proof
Assume that none of the $Y_k$ are empty.

If not, then any empty ones can simply be removed.

Uniqueness
Aiming for a contradiction, suppose that $Y_j, Y_k \in \mathcal U$ with $j \ne k$.

Then since $Y_1, \dots, Y_n$ are a pairwise disjoint:
 * $Y_j \cap Y_k = \varnothing$

But by the definition of an ultrafilter, $\mathcal U$ has the finite intersection property.

It follows from this contradiction that no such $Y_j, Y_k \in \mathcal U$ with $j \ne k$ exists.

Existence
Aiming for a contradiction, suppose that $Y_1, \dots, Y_n \notin \mathcal U$.

Since the $Y_k$ are pairwise disjoint:
 * $\displaystyle Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\}$

Then by the definition of an ultrafilter:
 * $\displaystyle \forall i: Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\} \in \mathcal U$

But:
 * $\displaystyle \bigcap_{i \mathop = 1}^n \bigcup \left\{{Y_k: k \ne i}\right\} = \varnothing$

contradicting the fact that $\mathcal U$ has the finite intersection property.

Thus $Y_k \in \mathcal U$ for some $k$.