Proof.

1. (1)

   Observe $(A\cap B)\inv=\curl{x\inv:x\in A\cap B}=\curl{x\inv:x\in A}\cap \curl{x\inv:x\in B}=A\inv\cap B\inv$ and $xA\cap yB\neq\emp\iff \exists a\in A,b\in B, xa=yb\iff \exists a\in A,b\in B, ab\inv=x\inv y\iff x\inv y\in AB\inv$.

2. (2)

   The first assertion is equivalent to continuity of multiplication and inversion respectively. The second assertion follows from the first.

3. (3)

   By continuity pick neighborhoods $W_{1},W_{2}$ of $1$ such that $W_{1}W_{2}\suf U$ then $V\define W_{1}\cap W_{2}\cap W_{1}\inv\cap W_{2}\inv$ then $V$ is symmetric by (1) and by construction $V\suf W_{1},W_{2}$ hence $VV\suf W_{1}W_{2}\suf U$.

4. (4)

   If $A,B$ are compact then $AB$ is compact by continuity of multiplication.

5. (5)

   For each $V\suf G$ we have $q\inv(q(V))=VH$ hence if $V$ is open then $q(V)$ is open by (2).

6. (6)
   1. (a)

      Let $x,y\in \cl{H}$ then there are nets $\curl{x_\al},\curl{y_\be}$ in $H$ converging to $x,y$ respectively so by continuity of group operations $xy,x\inv\in \cl{H}$

   2. (b)

      $G\del H=\cups{}{x\in G\del H}xH$ is open hence $H$ is closed.

   3. (c)

      Let $q(x)\neq q(y)\in G/H$ then $xHy\inv$ doesn’t contain 1 and closed by (2). By (3) there exists a symmetric neighborhood $U$ of $1$ such that $UU\cap xHy\inv=\emp$. hence $(UxH)(UyH)\inv=UxHy\inv U$ doesn’t contain $1$ so $(UxH)\cap (UyH)=\emp$ so $q(Ux),q(Uy)$ are disjoint neighborhoods of $q(x),q(y)$ respectively.

   4. (d)

      For each $x\in G$, conjugation by $x$, $c_{x}$ is continuous and by normality $c_{x}(\cl{H})=\cl{c_x(H)}=\cl{H}$. By normality $q(x)q(y)=xHyH=xyH=q(xy)$ for each $x,y\in G$. The result follows from (5) and Lemma 3.9.3(3).

   5. (e)

      If $U$ is a compact neighborhood of $1$ in $G$ then for any $x\in G$, $q(Ux)$ is a compact neighborhood of $q(x)$ in $G/H$.

   6. (f)

      Suppose $B$ is a neighborhood of $1$ then there exists $U\suf G$ open such that $1\in U\suf B$ then $AU$ is open and $A\suf AU\suf AB$.

   
   

$\square$