If $D \subseteq X$ is any subspace of $X$ that is homeomorphic to a closed set $E \subseteq Y$ and $Cl_Y(Int_Y(E)) = E$, then does $Cl_X(Int_X(D)) =D?$
I asked a very similar question to this in the past except with the added condition that $D$ was a closed set in $X$.
I know that since $D$ is homeomorphic to $E$, their interiors are homeomorphic, that is $Int(D)$ is homeomorphic to $Int(E)$.
On
Counter-example: $X=Y=[0,1]\cup [2,3)$ and $D=[0,1)$ and $E=[2,3).$
The function $f:D\to E,$ where $f(x)=x+2,$ is a homeomorphism.
Since $E$ is open and closed, $Cl(Int(E))=Cl(E)=E.$
But $Cl(Int(D))=Cl(D)=[0,1]\ne D.$
On
Counter-example: $X=Y=[0,1]\cup [2,3)$ and $D=[0,1)$ and $E=[2,3).$
Since $E$ is open and closed, $Cl(Int(E))=Cl(E)=E.$
But $Cl(Int(D))=Cl(D)=[0,1]\ne D.$
An even simpler counter-example: Let $X=[0,1]$ and $Y=[0,1)$ with $D=E=[0,1).$.
In general, let $T_Y$ be the topology on $Y,$ with $Y\ne \emptyset.$ Let $X=Y\cup \{p\}$ with $p\not \in Y.$ Let the topology on $X$ be $T_Y\cup \{t\cup \{p\}: \emptyset \ne t\in T_Y\}.$ Let $D=E=Y.$
Let $f$ be a homeomorphism from $X$ to $Y$. Indeed, as shown in this post, $f(\operatorname{Cl}(E))=\operatorname{Cl}(f(E))=\operatorname{Cl}(D).$ Therefore, as you noted that $f(\operatorname{Int}(E))=\operatorname{Int}(f(E))$, $$f(\operatorname{Cl}(\operatorname{Int}(E)))=\operatorname{Cl}(f(\operatorname{Int}(E)))=\operatorname{Cl}(\operatorname{Int}(f(E)))=\operatorname{Cl}(\operatorname{Int}(D))=f(E)=D.$$
If, however, $f$ is a homeomorphism from $E$ to $D$- and not necessarily from $X$ to $Y$-, things are different. Consider $[0,1)$ with the subspace topology of $\mathbb{R}$ and $(0,1]$ as a subset of $[0,1]$. Let, as a counter-example, $f:[0,1)\to[0,1]$ be an imbedding of $[0,1)$ to $(0,1]$. Indeed, $\mathrm{Cl}_{[0,1)}(\mathrm{Int}_{[0,1)}([0,1)))=[0,1)$, but $\mathrm{Cl}_{[0,1]}(\mathrm{Int}_{[0,1]}((0,1]))=\mathrm{Cl}_{[0,1]}((0,1])=[0,1]$.