It is stated in lots of places (including this one) that these two are equivalent definitons for an ultrafilter $u$ to be a p-point.
If for every sequence $\left < A_n: n \in \omega \right>$ of elements of $u$ there exists a $B$ in $u$ such that for all $n \in \omega$, $B \subseteq^* A_n$.
For every partition $\omega = \bigcup_{n \in \omega} A_n$ of $\mathbb{N}$ into sets that are not in $u$ ... ($A_n \notin u$), there is a $B \in u$ such that for every $n \in \omega$, $B\cap A_n$ is finite.
It says everywhere that this is easy, and indeed direction $(1)\rightarrow(2)$ is easy. However, I can't prove the other direction ...
Thanks.
Suppose the second definition, and let $\langle A_n\mid n\in\omega\rangle$ be a sequence in $u$. We may assume without loss of generality that the sequence is strictly decreasing, and that $A_0=\omega$. If the intersection is in $u$ we are done. So suppose not.
Consider now the partition defined by $B_n=A_n\setminus A_{n+1}$. We might have to add to this another part which is $\bigcap A_n$ if it is not empty. But now we have a partition of $\omega$ and none of the parts is in $u$ (since if $B_n\in u$ for some $n$ then we have $B_n\cap A_{n+1}=\varnothing\in u$). So there is some $B\in u$ which meets each of the $B_n$'s on a finite set only.
Now by induction, $B\subseteq A_0$ so of course $B\subseteq^* A_0$. Suppose that $B\subseteq^* A_n$. Now we have that $B\cap A_{n+1}=(B\cap A_n)\setminus B_n$, but since $B_n\cap B$ is finite we have that $B\cap A_n=^* B\cap A_{n+1}$, and so $B\subseteq^*A_{n+1}$.
Therefore the $A_n$'s have a pseudo intersection in $u$. $\qquad\square$