Component of Finite Union in Ultrafilter

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

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

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

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

Proof
Assume that none of the $Y_k$ are empty&mdash;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, a contradiction.

Existence
Suppose for the sake of contradiction that $Y_1, \dots, Y_n \notin \mathcal U$.

Since the $Y_k$ are pairwise disjoint, $Y_{i}^c = \bigcup \{ Y_k: k \ne i \}$.

Then by the definition of an ultrafilter, $Y_{i}^c = \bigcup \{ Y_k: k \ne i \} \in \mathcal U$ for each $i$.

But $\bigcap_{i \mathop = 1}^n \bigcup \{ Y_k: k \ne i \} = \varnothing$, contradicting the fact that $\mathcal U$ has the finite intersection property.

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