Non-existence of countable base of arbitrary ultrafilter

$$\scr{B}$$ is the base of a nonprincipal ultrafilter $$\scr{U}$$ on $$\omega$$ if 1. $$\forall U,V\in\mathscr{B}~\exists T\in\mathscr{B}:~T\subset U\cap V$$, 2. $$\forall X\in\mathscr{U}~\exists U\in\mathscr{B}: U\subset X$$. It is known that there is no countable base for nonprincipal ultrafilter. How we prove this?

• I think this question would be more appropriate at math.stackexchange. – Noah Schweber Apr 23 '19 at 13:25

Let $$\{U_i\}_{i<\omega}$$ be countable base and $$A\cap U_i\neq\varnothing$$ for all $$i<\omega$$. Then $$A\cap B\neq\varnothing$$ for all subsets $$B$$ from ultrafilter and thus $$A\in \scr{U}$$. But we can find distinct pairs $$\{a_i;b_i\}\in U_i$$ for all $$i$$ because $$U_i$$ are infinite. For the sets $$A=\{a_i\}$$ and $$B=\{b_i\}$$ we have $$A\cap U_i\neq\varnothing$$, $$B\cap U_i\neq\varnothing$$. So, $$A,B\in\scr{U}$$. But $$A\cap B=\varnothing$$. Contradiction.