(a) $x\cdot y$ and $x\cdot y^{-1}$ are continuous, therefore, there is a neighborhood $V'$ of $e$ such that $V'\cdot V'\subset U$ and $W$ such that $V=W\cdot W^{-1}\subset V'$ . Therefore, $V\cdot V\subset U$ , and $x\in V^{-1}$ iff $x=(y\cdot y^{-1})^{-1}=y\cdot y^{-1}$ for some $y\in W$ iff $x\in V$ .

(b) We need a symmetric neighborhood of $e$ such that for no $u^{-1},v\in V$ : $u^{-1}\cdot x=v\cdot y$ or $x\cdot y^{-1}=u\cdot v$ . In other words, we need a symmetric neighborhood $V$ of $e$ such that $V\cdot V$ does not contain $x\cdot y^{-1}$ . $G$ is a $T_{1}$ -space, therefore, $\{x\cdot y^{-1}\}$ is closed. Using (a), we construct $V$ for $U=G-\{x\cdot y^{-1}\}$ . It remains to note that since $g_{x}$ and $g_{y}$ are homeomorphisms (Exercise 4), $V\cdot x$ and $V\cdot y$ are disjoint open neighborhoods of $x$ and $y$ .

(c) We need a symmetric neighborhood of $e$ such that for no $u^{-1},v\in V$ and $y\in A$ : $u^{-1}\cdot y=v\cdot x$ or $y\cdot x^{-1}=u\cdot v$ . In other words, we need a symmetric neighborhood $V$ of $e$ such that $V\cdot V$ does not intersect $A\cdot x^{-1}$ . $A\cdot x^{-1}$ is closed ($g_{x^{-1}}$ is a homeomorphism by Exercise 4). Using (a), we construct $V$ for $U=G-A\cdot x^{-1}$ . It remains to note that since $g_{x}$ and $g_{y}$ for $y\in A$ are homeomorphisms (Exercise 4), $V\cdot x$ and $V\cdot A=\cup_{y\in A}V\cdot y$ are disjoint open neighborhoods of $x$ and $A$ .

(d) If $H$ is closed, $G/H$ is a $T_{1}$ -space (Exercise 5(b)). So, together with the regularity axiom we are about to prove, this shows that $G/H$ is a regular space. However, as noted in the comments below (see the comments by shafi and Daniel Whitman below), the regularity axiom holds even if $H$ is not closed (and, therefore, $G/H$ is not a $T_{1}-$ space).

A closed subset of $G/H$ is an image of a closed saturated subset of $G$ . Let $A$ be a saturated closed subset of $G$ that does not intersect $xH$ . Then, $A=A\cdot H$ . Indeed, $A=$ $p^{-1}(p(A))=$ $p^{-1}(\cup_{a\in A}aH)=$ $\cup_{a\in A}aH=$ $A\cdot H$ . Alternatively, you can just say that then $A$ consists of disjoint subsets $xH$ for some $x\in G$ each being closed under right multiplication by $H$ . As in (c), let $V$ be a symmetric neighborhood of $e$ such that $V\cdot A$ does not intersect $V\cdot x$ . Using Exercise 5(c), $p(V\cdot A)=\cup_{v\in V,a\in A}vaH$ is an open neighborhood of $p(A)$ disjoint from $p(V\cdot x)=\cup_{v'\in V}v'xH$ , an open neighborhood of $xH$ . Indeed, if they are not disjoint, then for some elements $v,v'\in V$ and $h,h'\in H$ , $v\cdot a\cdot h=v'\cdot x\cdot h'$ , and $v\cdot a\cdot h\cdot h'^{-1}=v'\cdot x$ where $a\cdot h\cdot h'^{-1}\in A\cdot H=A$ , contradicting $V\cdot A\cap V\cdot x=\emptyset$ .