Constructing a particular type of ultrafilter. Let $u$ be a free ultrafilter on $\omega$.
I am interested in constructing an ultrafilters on the closed subsets of $\mathbb R$ which contain the collection $\mathcal C$$=${$\bigcup_{n\in A} [n,n+1]:A\in u$}.
If $(x_n)\in \mathbb R ^\omega$ is a sequence of reals with the property $x_n\in (n,n+1)$ for each $n\in\omega$, then I can show the filter generated by $\mathcal C\cup${$x_n:n\in \omega$} is an ultrafilter.
Question: It is not true that every ultrafilter of the type I want can be constructed in this manner, but I am having some difficulty seeing why.  Could someone describe an ultrafilter containing $\mathcal C$ which can not be generated from $\mathcal C$ and a sequence (of the type I used)?
EDIT: What about this. For each $A\in u$, let $(s^A_n)_{n\in A}$ be a real-valued sequence on $A$. Then consider a collection $\mathcal S$$=${$(s^A_n):A\in u$} that has the finite intersection property. It looks like $\mathcal C\cup \mathcal S$  generates an ultrafilter?
 A: My original answer overlooked, as David pointed out in a comment, that he wants an ultrafilter in the lattice of closed sets, not in the Boolean algebra of all subsets of $\mathbb R$.  I'll append a correction below.  
First, here's the original, wrong answer:
Let $\mathcal B$ be the collection of those subsets of $\mathbb R$ whose intersections with the intervals $[n,n+1]$ have at most $1$ element (for each $n$). Then let 
$$
\mathcal D=\mathcal C\cup\{\mathbb R-X:X\in\mathcal B\}.
$$ 
This family $\mathcal D$ has the finite intersection property, so it can be extended to an ultrafilter on $\mathbb R$.  And this ultrafilter cannot contain any set of the form $\{x_n:n\in\omega\}$ as in your question, because sets of this form are in $\mathcal B$, so their complements are in $\mathcal D$.
Now here's the correction:
As David said, $\mathcal D$ should consist of closed sets, so $\mathcal B$ should consist of open sets, so I'll define a corrected version of $\mathcal B$.  Put an open subset $X$ of $\mathcal R$ into $\mathcal B$ if, for each $n$, the Lebesgue measure of $X\cap[n,n+1]$ is at most $2^{-n}$.  
Thie resulting corrected $\mathcal D$ still has the finite intersection property, because any set in $\mathcal C$ contains intervals $[n,n+1]$ for arbitrarily large $n$, and finitely many sets from $\mathcal B$ ccan't cover all such intervals because their total measure will be too small when $n$ is large.  So there is an ultrafilter of closed sets extending $\mathcal D$.
This ultrafilter cannot contain any set $\{x_n:n\in\omega\}$ as in the question because any such set is covered by a set in $\mathcal B$.
