Closure operators coincide means that the topologies coincide Assume there are two topologies on a single space such that  the closure of any given set in either topology is the same.
Is it true that the topologies are also the same? I think they are: assume there is an open set in the first topology that is not open in the second topology. Then, its complement in the first topology is closed, so the complement of the set is equal to its own closure in the first topology, which is equal to the complement’s closure in the second topology. Hence, the complement is closed in the second topology, which means the complement of the complement is open in the second topology. But the complement  of the complement of a set is the set itself. Hence the set is open in the second topology and we have a contradiction.
If this is false: what if we impose that the topological be sequential spaces?
 A: Your reasoning is correct.
Note that the closure operator defines which sets are closed. A closed set is exactly one whose closure is itself. So two topologies with the same closure operator have the same closed sets, hence the same open sets.
The same argument applies for the interior operator, which defines which sets are open. A set is open if and only if its interior is itself. So two topologies with the same interior operator have the same opens.
A: Let $X$ be any set and $\tau_1$ and $\tau_2$ be two topologies on $X$.
Then $\tau_1=\tau_2$ of $U\in \tau_1 \iff U\in \tau_2$
The following conditions are equivalent:

*

*$\tau_1=\tau_2$


*$ F\subset X$ is $\tau_1 $-closed iff $F$ is $\tau_2$ -closed.

Suppose $\forall A\subset X$ , $\operatorname{cl_{\tau_1}}(A)=\operatorname{cl_{\tau_2}}(A) $
Then $A\subset X$ $\tau_1$-closed implies $A=\operatorname{cl_{\tau_1}}(A)=\operatorname{cl_{\tau_2}}(A)$.Hence $A\subset X$ is also $\tau_2$ -closed and converse also true.
Note : Coincidence of closure implies $\text{Id}:(X,\tau_1)\to (X, \tau_2)$ is a homeomorphism.
Conclusion:
Coincidence of closure
$\iff$  coincidence of closed sets $\iff$  coincidence of open sets $\iff $ coincidence of interior $\iff$  coincidence of topologies.

Every closed set is sequentially closed but not the converse.
Question : Does coincidence of sequential closure implies coincidence of topologies?
Answer: NO.
For an example , let $X$ be any uncountable set.
$\tau_d :=\tau_{\text{dis}}=\mathcal{P}(X)$
$\begin{align}\tau&:=\tau_{\text{coc}}\\&
=\{U\subset X : U=\emptyset \text { or} |X\setminus U|\le \aleph_{0}\}\end{align}$
Then $(X, \tau_{\text{dis}}) $ and $(X, \tau_{\text{coc}}) $ has same collection of convergent sequences ( all eventually constant sequences).
Hence every subsets of $X$ in both topological spaces are closed.
See here.
