The closure of a subset in finer topology is always subset of the closure of that subset in the coarser one. On the Appendix A of Naber's book $\textit{The Geometry of Minkowski Spacetime}$ there is a claim in Lemma A.3.3. It says that if we have a set (says $M$) endowed with two different topology says $(M,O_A)$ and $(M,O_B)$ where $O_A$ finer than $O_B$ then for any subset $U$ of $M$ Closure $U$ in $O_A$ is always subset of Closure $U$ in $O_B$ ($Cl_A(U)$ $\subseteq$ $Cl_B(U)$).
I can show that this is true, but i 'm not sure my proof is valid.
Here's my proof. If you found this not valid or unsatisfying, please let me know.

The claim is that $Cl_A(U)$ $\subseteq$ $Cl_B(U)$. So, we must show that if $ x \in Cl_A(U)$, then $x \in Cl_B(U)$. This is equal to if $x \notin Cl_B(U)$ then $x \notin Cl_A(U)$.
$x \notin Cl_B(U) \iff x \in M-Cl_B(U)$. Because the closure of any subset is closed, then the complement must be open. So $M-Cl_B(U)$ open in $O_B$. But $O_A$ finer than $O_B$. So $M-Cl_B(U)$ must be open in $O_A$ too.
$x \in M-Cl_B(U)=M-(M-(Int_B(U) \cup Ext_B(U)) \iff x \in Int_B(U) \cup Ext_B(U) $
But $ Int_B(U) \cup Ext_B(U) $ is open in $O_A$.So x is in some open set in $O_A$. If $x \in Int_B(U) \implies x \in U$. But U may or may not open. So the largest open set that contain x is $Int_A(U)$. So $x \in Int_A(U)$. If $x \in Ext_B(U) \implies x \notin U$. Because x must contain in some open set in $O_A$ therefore $x \in Ext_A(U)$. So we have shown that
$x \notin Cl_B(U) \iff x \in Int_B(U) \cup Ext_B(U) $ which is implies $ x \in Int_A(U) \cup Ext_A(U) = M - (M-(Int_A(U) \cup Ext_A(U)))=M-Cl_A(U) \iff x \notin Cl_A(U)$.


Easier proof: $Cl_B(U)$ is an $A$-closed set which contains $U$, hence $Cl_B(U)$ must also contain $Cl_A(U)$, by definition of $Cl_A$.
