I start with the set $$ X=\{a,b,c,d\} $$ and the subbasis $$ \{\{\},\{a,b\},\{a,d\},\{b,c,d\},\{a,b,c,d\}\} $$
The I generate the topology by first taking all possible (finite) intersections: $$ \{\{\},\{a\},\{b\},\{a,b\},\{d\},\{a,d\},\{b,c,d\},\{a,b,c,d\}\} $$ and then all union, obtaining the topology to the subbasis: $$ \{\{\},\{a\},\{b\},\{a,b\},\{d\},\{a,d\},\{a,b,d\},\{b,c,d\},\{a,b,c,d\}\} $$
The the closed set by taking all complements: $$ \{\{\},\{a\},\{c\},\{b,c\},\{a,b,c\},\{c,d\},\{a,c,d\},\{b,c,d\},\{a,b,c,d\}\} $$
Then I list all closures obtained by taking the intersection of all closed supersets of some given set: $$ \text{cl}\{a\}=\{a\},\\ \text{cl}\{b\}=\{b,c\},\\ \text{cl}\{a,b\}=\{a,b,c\},\\ \text{cl}\{c\}=\{c\},\\ \text{cl}\{a,c\}=\{a,c\},\\ \text{cl}\{b,c\}=\{b,c\},\\ \text{cl}\{a,b,c\}=\{a,b,c\},\\ \text{cl}\{d\}=\{c,d\},\\ \text{cl}\{a,d\}=\{a,c,d\},\\ \text{cl}\{b,d\}=\{b,c,d\},\\ \text{cl}\{a,b,d\}=\{a,b,c,d\},\\ \text{cl}\{c,d\}=\{c,d\},\\ \text{cl}\{a,c,d\}=\{a,c,d\},\\ \text{cl}\{b,c,d\}=\{b,c,d\} $$
The problem is with the set $\{a,c\}$: (i) its complement is not open ($\{b,d\}$ is not in the topology); (ii) the closed super set of $\{a,c\}$ are $\{a,b,c\},\{a,c,d\},\{a,b,c,d\}$ whose intersection is $\{a,c\}$. That is, $\{a,c\}$ is not closed but $$ \text{cl}\{a,c\}=\{a,c\} $$ a contradiction.
There must be an error. Where?
The singletons $\{b\}$ and $\{d\}$ are in your collection of open sets, but you left out the set $\{b,d\}$. Its complement is $\{a,c\}$.