Could someone please check my work against the definition and Theorem provided

Definition : Ler s be a subset of R. A point x in R is an inferior point of S if there exists a neighborhood Nofx such That NSS. Theorem : A ser S is open iff S= inTS. Equivalently, S is open iff every point in S is an inferior point of S. Ler S and T be subsers of IR. Prove: inT (S) is an open set using The definition and Theorem above. Show That Hint: Vx E inT (S ) FE>0 such That N( x; E) E inT (S). x-a 6 -x Ler int ( s ) = ( 2, 6 ). Vab ER Lef XE inT (S ) Ler E = Min {x-a, 6-x3 => N ( x ; E) E int (S) by definition Since x is arbitrary, > any x is an interior point of int (S ). By The Theorem , Since every x is an interior point of int (S ), Then int (S ) is open. Thus, inT (s ) is an open ser