This is essentially a question from Engelking's text.
Suppose that $( X , \mathcal{O} )$ is a (Hausdorff) space, and let $\mathcal{O}^\prime$ be the topology on $X$ generated by the family of all regular open subsets of $( X , \mathcal{O} )$. Show that the regular open subsets of $( X , \mathcal{O}^\prime )$ are precisely the regular open subsets of $( X , \mathcal{O} )$.
This is connected to the title because a topological space $X$ is semiregular if (it is Hausdorff and) the family of all regular open subsets of $X$ forms a base for $X$. Recall that $A \subseteq X$ is regular open if $\operatorname{Int} ( \overline{A} ) = A$.
So far I've only really been able to figure out the trivial observations. Clearly $\mathcal{O}^\prime$ is coarser than $\mathcal{O}$, and therefore $$ \operatorname{Int}_{\mathcal{O}^\prime} (A) \subseteq \operatorname{Int}_{\mathcal{O}} (A) \subseteq A \subseteq \operatorname{cl}_{\mathcal{O}} (A) \subseteq \operatorname{cl}_{\mathcal{O}^\prime} (A). $$ From this I can build a table showing the basic relationships between $A$, $\operatorname{cl}_{\mathcal{O}} (A)$, $\operatorname{cl}_{\mathcal{O}^\prime} (A)$, $\operatorname{Int}_{\mathcal{O}} (A)$, $\operatorname{Int}_{\mathcal{O}} ( \operatorname{cl}_{\mathcal{O}} (A))$, $\operatorname{Int}_{\mathcal{O}} ( \operatorname{cl}_{\mathcal{O}^\prime} (A))$, $\operatorname{Int}_{\mathcal{O}^\prime} (A)$, $\operatorname{Int}_{\mathcal{O}^\prime} ( \operatorname{cl}_{\mathcal{O}} (A))$, $\operatorname{Int}_{\mathcal{O}^\prime} ( \operatorname{cl}_{\mathcal{O}^\prime} (A))$, but as it hasn't seemed to help, I think I'll omit it.
After this I've got nothing.
I have a feeling that your table of relationships between those sets was missing at least one equality!
Let's begin with a small (but apparently central) fact about this situation:
Fact. Given an $\mathcal{O}^\prime$-open $\newcommand{\newcl}{\operatorname{cl}_{\mathcal{O}^\prime}}\newcommand{\oldcl}{\operatorname{cl}_{\mathcal{O}}}\newcommand{\newint}{\operatorname{Int}_{\mathcal{O}^\prime}}\newcommand{\oldint}{\operatorname{Int}_{\mathcal{O}}}U \subseteq X$, we have that $\newcl (U) = \oldcl (U)$.
proof. As noted in the question, we already have $\oldcl(U) \subseteq \newcl (U)$. Suppose that $x \in X \setminus \oldcl (U)$. Setting $V := X \setminus \oldcl (U)$, clearly $V \cap U = \varnothing$. Since $U$ is also $\mathcal{O}$-open it follows that $\oldcl (V) \cap U = \varnothing$, and thus $\oldint ( \oldcl (V) ) \cap U = \varnothing$. Since $\oldint ( \oldcl (V) )$ is $\mathcal{O}$-regular-open (the interior of any closed set is regular open), it is $\mathcal{O}^\prime$-open. As $x \in V \subseteq \oldint ( \oldcl (V) )$, it follows that $x \notin \newcl(U)$. $\quad\Box$
Using the duality between the interior and closure operators (i.e., $\operatorname{Int} (A) = X \setminus \overline{ X \setminus A }$), we get the following corollary.
Coro. Given an $\mathcal{O}^\prime$-closed $F \subseteq X$, we have that $\newint (F) = \oldint (F)$.
Suppose that $U \subseteq X$ is $\mathcal{O}$-regular-open. Note that $U$ is $\mathcal{O}^\prime$-open, and so by the fact above, $\newcl (U) = \oldcl (U)$. Thus, $$U \subseteq \newint ( \newcl (U) ) = \newint ( \oldcl (U) ) \subseteq \oldint ( \oldcl (U) ) = U,$$ and so $\newint ( \newcl (U) ) = U$, meaning that $U$ is $\mathcal{O}^\prime$-regular-open.
Suppose that $U \subseteq X$ is $\mathcal{O}^\prime$-regular-open. Since $\newcl(U)$ is $\mathcal{O}^\prime$-closed, by the corollary above we know that $\newint(\newcl(U)) = \oldint(\newcl(U))$, and so $$U \subseteq \oldint ( \oldcl (U) ) \subseteq \oldint ( \newcl (U) ) = \newint ( \newcl (U) ) = U,$$ and so $\oldint ( \oldcl (U) ) = U$, meaning that $U$ is $\mathcal{O}$-regular-open.