We need to check that $C$ satisfies the defining properties.

Property 1: Since every element of the set $\{Y\in K\mid X\subseteq Y\}$ contains $X$, we have $X\subseteq C(X)$.

Property 2: For every element $Y$ of $K$ such that $X\subseteq Y$, it also is the case that $C(X)\subseteq Y$ because an intersection of a family of sets is a subset of any member of the family. In other words (or rather, symbols),

hence $C(C(X))\subseteq C(X)$. By the first property proven above, $C(X)\subseteq C(C(X))$ so $C(C(X))=C(X)$. Thus, $C\circ C=C$.

Property 3: Let $X$ and $Y$ be two subsets of $L$ such that $X\subseteq Y$. Then if, for some other subset $Z$ of $L$, we have $Y\subset Z$, it follows that $X\subset Z$. Hence,

so $C(X)\subseteq C(Y)$.

∎