reserve X for non empty TopSpace;

theorem
  for A being Subset of X holds
  Int A = union {G where G is Subset of X : G is open & G c= A}
proof
  let A be Subset of X;
  set F = {G where G is Subset of X : G is open & G c= A};
A1: F c= bool the carrier of X
  proof
    let C be object;
    assume C in F;
    then ex P being Subset of X st C = P & P is open & P c= A;
    hence thesis;
  end;
  {} c= A;
  then {}X in F;
  then reconsider F as non empty Subset-Family of X by A1;
  now
    let P be set;
    assume P in F;
    then ex G being Subset of X st G = P & G is open & G c= A;
    hence P c= A;
  end;
  then
A2: union F c= A by ZFMISC_1:76;
  Int A c= A by TOPS_1:16;
  then Int A in F;
  then
A3: Int A c= union F by ZFMISC_1:74;
  now
    let S be Subset of X;
    assume S in F;
    then ex G being Subset of X st G = S & G is open & G c= A;
    hence S is open;
  end;
  then F is open by TOPS_2:def 1;
  then union F c= Int A by A2,TOPS_1:24,TOPS_2:19;
  hence thesis by A3;
end;
