Regular open/closed set Prove that the
i) int(A) is a regular open set for every closed set A
ii) closure(U) is a regular close set for every open set U
For i) is this a valid solution. 
We have to show that int(A)=int[closure(intA)]
Since A is closed set, then A=closure[int(A)] then int(A)=int(A) ???
 A: $\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Your argument for (i) doesn’t work, because it’s not true in general that $A=\cl\int A$ when $A$ is a closed set. For example, suppose that $A$ is the middle-thirds Cantor set in $\Bbb R$; then $A$ is closed, but $\int A=\varnothing$, so $\cl\int A=\varnothing\ne A$. For that matter, just let $A=\{0\}$: this is also a closed set in $\Bbb R$, but $\cl\int\{0\}=\cl\varnothing=\varnothing\ne\{0\}$.
Let $A$ be a closed set in a space $X$, and let $U=\int A$. Clearly $U\subseteq A$, so $\cl U\subseteq\cl A=A$, and therefore $\int\cl U\subseteq\int A=U$. It only remains to show that $U\subseteq\int\cl U$. HINT: Start from the fact inclusion $U\subseteq\cl U$ and take interiors on both sides.
For (ii) you can argue in very similar fashion. Start with an open set $U$, and let $A=\cl U$. Then $U\subseteq A$, so $U=\int U\subseteq\int A$, and ... ? 
Alternatively, you can start with an open set $U$ and let $A=X\setminus U$. Then $A$ is closed, so by (i) we know that $\int A$ is regular open. If you know that the complement of a regular open set is regular closed, you can conclude that $X\setminus\int A$ is regular closed and try to show that it’s equal to $\cl U$.
A: For every open set $O$ we have $O=\text{int}(O)$ and thus $O\subseteq \text{int}(\overline O)$ (Why?)
Similarly, if $A$ is closed, we have $A=\overline A$, thus also $A \supseteq\overline{\text{int}(A)}$.
There is one special property of the closure and interior operators, that you should use throughout the proof.
A: If A= Int cl A , A is regular
If we can prove that A ⊆ Int cl A and Int cl A ⊆ A then A is regular set.
We know that A ⊆ Int cl A 
and Int A  ⊆  A and A is closed A = cl A, then Int cl A ⊆ A
therefore, A = Int cl A, A is regular
In same way we can prove  2
