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Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega_x$. Now take the set of all sets $Y$ that intersect all but finitely many vertical lines by a subset from $\omega_x$ (and don't intersect the other lines at all). Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega_x$. Thus $\omega$ is a counterexample.

Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega_x$. Now take the set of all sets $Y$ that intersect all but finitely many vertical lines by a subset from $\omega_x$. Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega_x$. Thus $\omega$ is a counterexample.

Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega_x$. Now take the set of all sets $Y$ that intersect each vertical line by a subset from $\omega_x$. Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega_x$. Thus $\omega$ is a counterexample.

Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega_x$. Now take the set of all sets $Y$ that intersect all but finitely many vertical lines by a subset from $\omega_x$ (and don't intersect the other lines at all). Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega_x$. Thus $\omega$ is a counterexample.

Bjorn's answer is not quite correct because the quantifier for $C$ is $\exists$. But it can be fixed. Take a partition of $\mathbb N$ into 3-element subsets $\{3n+1,3n+2, 3n+3\}$, $n=0,1,...$. Then let $C$ be the set of numbers not divisible by 3. Take a non-principle ultrafilter containing $C$ (it exists by the Zorn lemma). It is a counterexample: none of the partition classes is in the ultrafilter, because it is non-principle, and $C$ intersects each partition class by 2 elements.

Take a partition of ${\mathbb N}^2$ into vertical lines $\{x\}\times {\mathbb N}$. In each vertical line take a non-principle ultrafilter $\omega_x$. Now take the set of all sets $Y$ that intersect each vertical line by a subset from $\omega_x$. Note that all complements of finite sets of ${\mathbb N}^2$ are in our set of sets, and that it is clearly a filter. Take any ultrafilter $\omega$ that contains that filter. It exists by the Zorn lemma. Clearly, the first option does not hold: none of the vertical lines is in $\omega$. Now suppose that for some $C\in\omega$, $C$ intersects each vertical line by at most 1 element. Then its complement intersects each vertical line by a subset that is either the whole line or the line without one element. That is impossible because we chose non-principle $\omega_x$. Thus $\omega$ is a counterexample.