$A\in \tau_1$ ve $x\in A$ olsun.
$$\left.\begin{array}{rr} x\in A\in \tau_1 \\ \\ U:=A\end{array}\right\}\Rightarrow (U\in\mathcal{U}(x))(|U\setminus A|=|A\setminus A|=|\emptyset|=0\leq \aleph_0)$$ olduğundan $$A\in\tau_2$$ olur yani $$\tau_1\subseteq \tau_2.$$