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 $X-S_i \in F$.  So $\bigcap_{i \le n}(X-S_i) \in F$.  But $\bigcap_{i \le n}(X-S_i) = \emptyset \notin F$.
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, $X^c$ = $\bigcap_{i \le n}(X-Si) \in F$, but $X^c = \emptyset \notin F$, creating a contradiction.  Did I misunderstand something?

If I were writing such text,  I would have pointed it out in order to remind the reader of this. Especially when in just one sentence we derived a contradiction from it. 
