$H$ closed implies $G / H$ Hausdorff

Question

Let $G$ be a topological group and $H$ be a subgroup of $G$. It is to show that the closedness of $H$ implies $G/H$ being a Hausdorff space.

We denote the projection map $G \to G/H$ with $p$.

My book gives the following proof:

Let H be closed. Then, the preimage of H under the continuous map $$G \times G \to G, (g, g') \mapsto g^{-1}g'$$ is closed too. This is the set of pairs $(g, g')$ with $gH = g'H$, i. e. the equivalence relation belnging to the surjection $p$. Since $p$ is also open, the equivalence relation's closedness implies that the quotient is a Hausdorff space.

I do not understand the last sentence. Given two different points in $G/H$, how to find concretely disjoint neighbourhoods in $G/H$ and why do we need the function, since its domain is a product, not G itself? Why is $p$ open and how do we can conclude from this the Hausdorff property?

I have found a similar question here, but the way of proof was rather different. And yes, it's embarrassing not being able to understand a detailed simple proof. :-)

Note: On Wikipedia, I found that (applied to this situation) $ker(p)$ closed implies $G/H$ Hausdorff. This would ease the proof dramatically, but my book did not prove this fact.

The projection map. I edited the question to include that piece of information. — user7427029, Oct 15 '20 at 10:03
Looks like your textbook is using the following: If $\sim$ is an open equivalence relation, then $X/{\sim}$ is Hausdorff iff ${(x,y)\mid x\sim y} $ is closed in $X\times X $. — cqfd, Oct 15 '20 at 10:17
This has a proof of the above and more https://mathstrek.blog/2013/03/23/topology-quotients-of-topological-groups/ — abhi01nat, Oct 15 '20 at 10:42
The quotient map of a group action is open: see the comments to this question — YCor, Oct 18 '20 at 21:49

$H$ closed implies $G / H$ Hausdorff

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