Prove the Following Property of an Ultrafilter

Prove the Following Property of an Ultrafilter

In her text Introduction to Modern Set Theory , Judith Roitman defined a
filter of a set $X$ as a family $F$ of subsets of $X$ so that:
(a) If $A \in F$ and $X \supseteq B \supseteq A$ then $B \in F$.
(b) If $A_1, ... ,A_n$ are elements of $F$, so is $A1 \cap ... \cap An$.
Then she proceeded to define an ultrafilter as such: "If $F$ is proper
and, for all $A_n \subseteq X$,either $A \in F$ or $X-A \in F$, we say
that $F$ is an ultrafilter."
Now, suppose that $F$ is an ultrafilter on a set $X$. Prove that if $X =
S_1 \cup ... \cup S_n$, then some $S_n \in F$. She wrote, "If not, then,
since no $S_i \in F$, $F$ is proper, and each $X-Si \in F$. So $\cap_{i
\le n}(X-Si) \in F$. But $\cap_{i \le n}(X-Si) = \emptyset \notin F$.
What I did not understand was that if she already defined an ultrafilter
as proper, why did she have to say "since no $S_i \in F$, $F$ is proper
..."? My thinking was that if $X = S_1 \cup ... \cup S_n$ is not an
element of $F$, then by the fact that $F$ is an ultrafilter, $X^c$ =
$\cap_{i \le n}(X-Si) \in F$, but $X^c = \emptyset \notin F$, creating a
contradiction. Did I misunderstand something?