reserve GX for TopSpace;
reserve A, B, C for Subset of GX;
reserve TS for TopStruct;
reserve K, K1, L, L1 for Subset of TS;

theorem
  [#]GX \ A = B \/ C & B,C are_separated & A is closed implies A \/ B is
  closed & A \/ C is closed
proof
  assume that
A1: [#]GX \ A = B \/ C and
A2: B,C are_separated and
A3: A is closed;
  now
    let A,B,C be Subset of GX;
    assume that
A4: [#]GX \ A = B \/ C and
A5: B,C are_separated and
A6: A is closed;
A7: Cl A = A by A6,PRE_TOPC:22;
    (Cl B) misses C by A5;
    then
A8: (Cl B) /\ C = {};
A9: [#]GX = A \/ (B \/ C) by A4,XBOOLE_1:45;
    Cl(A \/ B) = (Cl A) \/ Cl B by PRE_TOPC:20
      .= A \/ ((Cl B) /\ (A \/ (B \/ C))) by A7,A9,XBOOLE_1:28
      .= (A \/ Cl B) /\ (A \/ (A \/ (B \/ C))) by XBOOLE_1:24
      .= (A \/ Cl B) /\ ((A \/ A) \/ (B \/ C)) by XBOOLE_1:4
      .= A \/ ((Cl B) /\ (B \/ C)) by XBOOLE_1:24
      .= A \/ (((Cl B) /\ B) \/ ((Cl B) /\ C)) by XBOOLE_1:23
      .= A \/ B by A8,PRE_TOPC:18,XBOOLE_1:28;
    hence A \/ B is closed by PRE_TOPC:22;
  end;
  hence thesis by A1,A2,A3;
end;
