TITLE: How to show that a free ultrafilter cannot have an infinite pseudointersection?
QUESTION [6 upvotes]: The following text is a quote from p.180 of Halbeisen's book Combinatorial Set Theory.
This book is also available  on website of a course taught by the author. (As mentioned in Asaf's comment, it is also available on the author's website.)

For two sets $x,y\subseteq\omega$ we say that $x$ is almost contained in $y$, denoted $x\subseteq^*y$, if $x\setminus y$ finite.
A pseudo-intersection of a family $\mathscr F \subseteq [\omega]^\omega$  of infinite subsets of $\omega$ is an infinite subset of $\omega$ that is almost contained in every member of $\mathscr F$.
Furthermore, a family $\mathscr F \subseteq [\omega]^\omega$  has the strong finite intersection property (sfip) if every finite subfamily has infinite intersection.
For example any filter $\mathscr F \subseteq [\omega]^\omega$  has the sfip, but no ultrafilter on $[\omega]^\omega$ has a pseudo-intersection.


How can we show that a free ultrafilter cannot have an infinite pseudointersection?

This fact is used to show that $\mathfrak p$ is well-defined and $\mathfrak p \le \mathfrak c$,
where 
the pseudo-intersection number $\mathfrak p$ is the smallest cardinality
of any family $\mathscr F \subseteq [\omega]^\omega$ which has the sfip but which does not have a pseudo-intersection.
I will post my proof below; but I wonder whether there are different ways to show this.

REPLY [5 votes]: Let $\mathcal U$ be an ultrafilter on $\omega$, and let $x$ be a pseudo-intersection of $\mathcal U$. Furthermore, let $y$ be such that both it and its compliment contain infinitely many elements of $x$ (for instance, $y$ could be your $B$). Then, either $y$ or $\omega\backslash y$ are in $\mathcal U$, so either $x\backslash y$ or $x\backslash (\omega\backslash y) = x\cap y$ is finite, which is impossible.