Proof that no σ-complete non-principal filter on a countable set can exist

Jech (3rd edition, page 73) defines a principal filter as {X : Xo ⊆ X} with Xo a non empty subset of S and (page 77) indicates that there is no non-principal σ-complete (i.e. ω1 complete) filter on a countable set S, but provides no proof, so I was hoping that someone would be able to provide a reference to its proof.

I have seen a principal filter defined differently - with Xo being replaced by a single element of S. In this case if S is ω and the filter is ultra, it can't contain any finite subsets of ω. The filter must therefore contain all co-finite subsets of ω and therefore all end segments of ω, whose countable intersection is the empty set. So there is no non-principal σ-complete (ie ω1 complete) ultra-filter on ω.

As Jech strengthens the definition of "principal" from a single element of S to a subset Xo of S, it looks like the ultra-filter requirement (which ensures that any finite subset of S in an ultra-filter is always associated with a single element of S being in the ultra-filter) could be dropped, which Jech statement above suggests is the case.

Using Jech principal filter definition, an ω1 complete non-principal filter on ω must have > |ω| elements, as otherwise it would be principal. But I can't see how this ensures there is always an empty intersection of a countable number of elements in the filter, without getting into Continuum Hypothesis territory.

If a set is countable we can write it as $A=\{a_n\mid n<\omega\}$. If $U$ is a non-principal filter on $A$, then for some $n$'s, $A\setminus\{a_n\}\in U$. Look at $B=\bigcap\{A\setminus\{a_n\}\mid A\setminus\{a_n\}\in U\}$
Now apply $\sigma$-closure and examine what happens when $B\in U$.