Simple question regarding closure and subsets, with condition on interior I recently asked this question and got a very quick answer. I realize now that I should have included an extra condition. I have included the new condition in the modified version of the question below.

Suppose I have two compact sets $A$ and $B$ and denote their interiors as int$(A)$ and int$(B)$. Suppose, additionally, that I know that $A=\overline{\text{int}(A)}$ and $B=\overline{\text{int}(B)}$ where $\overline{C}$ represents the closure of $C$.
Are the two below statements equivalent?
Statement $1$: $\text{int}(A) \subseteq \text{int}(B)$
Statement $2$: $A \subseteq B$
 A: Two facts here are needed. The first is that $int$ preserves the subset relation - that is, $X \subseteq Y$ implies $int(X) \subseteq int(Y)$. This follows from the relation $U \subseteq int(X)$ iff $U \subseteq X$ for all open $U$, since we have $U \subseteq int(X)$ implies $U \subseteq X$ implies $U \subseteq Y$ implies $U \subseteq int(Y)$, hence $int(X) \subseteq int(Y)$ (taking $U = int(X)$).
From the above fact, (2) clearly implies (1).
Similarly, closure preserves the subset relation. This follows from the relation $\overline{X} \subseteq C$ iff $X \subseteq C$ for all closed $C$. From this, if $X \subseteq Y$ then for all closed $C$, then for all closed $C$, $\overline{Y} \subseteq C$ implies $Y \subseteq C$ implies $X \subseteq C$ implies $\overline{X} \subseteq C$; hence, $\overline{X} \subseteq \overline{Y}$ since $\overline{Y}$ is closed.
From the above fact, (1) implies (2). For if $int(A) \subseteq int(B)$, then $A = \overline{int(A)} \subseteq \overline{int(B)} = B$.
A: If $\overline{(int A)} = A$ then for any $x \in A$ so that $x \not \in int A$ then $x$ is a limit point of $int A$. (and thus also a limit point of $A$).
In order for $int(A)\subset int (B)$ but $A \not \subset B$
there must be a point $x\in A$ where $x \not \in B$.  For that to be true we can't have $x \in int A$ (or else $x \in int (B)\subset B$) so we must have $x$ is a limit point of $int A$ (but is not an interior point of $A$) but is not a point of $B$ (thus not a limit point of $B$ [as $B$ is closed], nor a limit point of $int B$, nor an interior point of $B$).
So $x$ is a limit point of $int A$ so for any $r>0$ then $N_r(x)$ will contain an interior point point of $a'\in int A$.  And thus there is $d$ so that $N_d(a')\subset int A\subset int B$. But that means $a'\in int B$ so $x$ is a limit point of $\int B$  and therefore $x$ IS a point of $\overline{int B}=B$ after all!.
So...  $int(A) \subset int B \implies A \subset B$.
In order for $A \subset B$ but $int A \not \subset B$
there must be a point $x \in int A$ but $x \not \in int B$.
But that's easily shown to be impossible.  If $x\in int A$ then there exists a $d > 0$ so that $N_r(x)\subset A \subset B$ so $x$ is an interior point of $B$.
......
So yes, the statements are equivalent...
!!!IF!!! $A=\overline{int A}$ and $B=\overline{int B}$.
Note if $A = [0,2]\cup \{3,4,5\}$ then ${int A} = (0,2)$ and $\overline{int A} = [0,2] \ne A$ and if we have $B = [0,2]$ then......
......
However $A\subset B \implies int(A) \subset int (B)$ regardless of anything else (not the argument is just if $x\in int(A)$ then there is a neigborhood of $x$ entirely withing $A$ and thus entirely within $B$-- nothing other than $A \subset B$ was used)
And if $C \subset D$  then $\overline C \subset \overline D$ is always true (because if $x$ is a limit point of $C$ then every neighborhood of $x$ contains a point of $C$ with must also be a point of $D$ so $x$ is also a limit point of $D$).
Those are very handy and important results.  But if a set has weird pesky points that are neither interior nor limit points then the throw this off.  (But those points can usually be brushed off like coffee grounds as unimportant.)
