If $\tau_1\subset\tau_2$ then $K_2(A)\subset K_1(A)$
Solution 1:
Because $\tau_1 \subseteq \tau_2$ we know that if a set $C$ is $\tau_1$-closed it is $\tau_2$-closed as well. (then $X\setminus C \in \tau_1$ so $X\setminus C \in \tau_2$ so $C$ is $\tau_2$-closed.) Now:
$$K_2(A) = \bigcap \{C\mid C \text{ is } \tau_2\text{-closed} ; A \subseteq C\} \subseteq \bigcap \{C\mid C \text{ is }\tau_1\text{-closed} ; A \subseteq C\} = K_1(A)$$
as intersections of larger families only can get smaller.