$Int(\overline{U\cap M})=Int(\overline{U})\cap Int(\overline{M})$ for any U open in X Prove that $Int(\overline{U\cap M})=Int(\overline{U})\cap Int(\overline{M})$ for any U open in X 
1) To prove that $Int(\overline{U\cap M}) \subseteq Int(\overline{U})\cap Int(\overline{M})$ I used that $\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}$ and $Int(A\cap B)=Int(A)\cap Int(B) $ for any $A,B\subseteq X$.
Since $\overline{U\cap M}\subseteq\overline{U}\cap\overline{M},$ we get that $Int(\overline{U\cap M}) \subseteq Int(\overline{U}\cap\overline{M})=Int(\overline{U})\cap Int(\overline{M}) $
2)To prove the inverse inclusion,  $Int(\overline{U\cap M}) \supseteq Int(\overline{U})\cap Int(\overline{M})$ I tried to use that a point $x\in Int(A)\iff\exists U\text{ open in }X \text{ such that }x\in U\subseteq A$.
Let $x\in Int(\overline{U})\cap Int(\overline{M}).$ So $x\in Int(\overline{U})\text{ and }x\in Int(\overline{M}).$ There exist two neighbourhoods $W_{1}, W_{2}$ of $x$ such that $x\in W_{1}\subseteq \overline{U} \text{ and } x\in W_{2}\subseteq \overline{M}$. So $x\in W_{1}\cap W_{2}\subseteq \overline{U}\cap\overline M \text{ and thus }x\in  Int(\overline{U\cap M}) $ ( since $W_{1}\cap W_{2}$ is open).
I probably made a mistake somewhere in the second part since I never used that U is open, but I can't find it (I have seen a counterexample why it doesnt hold if U isn't open).
 A: int A is interior of A;  cl A is closure of A.
A /\ B is A intersection B;  A \/ B is A union B
This is part of an efficient plain text language in use
before the inefficient Math Jaxov monopolized math talk.  
Lemma.  If U is open, then U /\ cl A subset cl (U /\ A).
This is a very useful lemma that allows 'algebraic' proofs
instead of point proofs as you were attempting.  A point
proof, which I leave for you, is needed to prove the lemma.  
open A implies int cl A/\B = int cl A /\ int cl B
Proof.  Uses the lemma twice.  
int cl A/\B subset int (cl A /\ cl B) = int cl A /\ int cl B
subset int (cl A /\ int cl B) subset int cl (A /\ int cl  B)
subset int cl (A /\ cl B) subset int cl cl A/\B = int cl A/\B  
Immediately, by complementation, one has the topological dual
closed A implies cl int A\/B = cl int A \/ cl int B.  
A is regular open when A = int cl A.
Exercise.  Show int cl A is regular open and the
intersection of two regular open sets is regular open.  
Note that in your problem all the sets involved are regular
open.  However the problem, though similar, isn't as simple
as the intersection of regular open sets is regular open.  
