Open normal subgroups of a profinite group form a basis for the open neighborhoods of 1

Question

In the following link: An equivalent definition of the profinite group
I'm having trouble understanding the following quote from the answer:

So if you replace each subgroup in a basis by the intersection of its conjugates, you obtain a basis made up of normal subgroups.

How can one prove that these normal subgroups form a basis?

Answer 1

If $\mathscr H$ is a local base of $1$, then $\{\bigcap_{x\in G} xHx^{-1}\mid H\in \mathscr H\}$ is a local base of $1$ because for every open neighborhood $A$ of $1$, $1\in \bigcap_{x\in G} xHx^{-1}\subset H\subset A$ for some $H\in \mathscr H$. The fact that $\bigcap_{x\in G} xHx^{-1}\mid H$ is open is proven in your linked answer.