You should mean $\{x\}=\bigcap\xi$, and the answer is clearly yes, since we can take $\xi$ equal to the set of all open sets containing $x$. Any point $y$ other than $x$ is excluded in this intersection by the open set $X-\{y\}$. The cardinality of this $\xi$ is the same as the number of finite subsets of $X$, which is equinumerous with $X$. And the same will be true of any $\xi$ having your property.

You should mean $\{x\}=\bigcap\xi$, and the answer is clearly yes, since we can take $\xi$ equal to the set of all open sets containing $x$. Any point $y$ other than $x$ is excluded in this intersection by the open set $X-\{y\}$. The cardinality of this $\xi$ is the same as the number of finite subsets of $X$, which is equinumerous with $X$. And the same will be true of any $\xi$ having your property, since the points other than $x$ must get excluded by elements of $\xi$, but only finitely many at a time, and so the cardinality of $\xi$ must be the same as $X$.

(This argument uses the axiom of choice, in order to know that the collection of finite subsets of an infinite set is equinumerous with that set. In $\neg$AC worlds, the situation is more complicated.)