Could someone please check my work using included definition and theorem

Definition: Ler's be a subser of R. A point x in R. is an inferior point of S if there exists a neighborhood Nofx suck That NSS. Theorem : A sar S is open iff S= inTS. Equivalently, S is open iff every point in S is an interior 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 ERR a Let 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