Theorem Let $L$ be a subextension of the Galois extension $K/F$. Then $\text{Gal}(K/L)$ is a closed subgroup of $\text{Gal}(K/F)$. For simplicity, we name $H:=\text{Gal}(K/L)$ and $L:=K^H$. Have maps $$ H\iff L $$ which give an inclusion-reversing bijection between subfields $F\subset L\subset K$ and closed subgroups $H\subset G$. A subextension $L/F$ is Galois over $F\iff \text{Gal}(K/L)\trianglelefteq\text{Gal} (K/F)$; in this case there is a natural isomorphism $\text{Gal}(L/F) \simeq \text{Gal} (K/F)/\text{Gal} (K/L)$.
I'm having trouble understanding the converse part of the proof given in Tamas Szamuely's Galois groups and fundamental groups. Here is the portion I don't understand:
proof ...(forward direction omitted)...Conversely, given a closed subgroup $H\subset G$, it fixes some extension $L/F$ and is thus contained in $\text{Gal} (K /L)$. To show equality, let $\sigma\in\text{Gal}(K/L)$, and pick a fundamental open neighbourhood $U_M$ of the identity in Gal$(K/L)$, corresponding to a Galois extension $M/L$. Now $H\subset \text{Gal} (K/L)$ surjects onto $\text{Gal} (M /L)$ by the natural projection; indeed, otherwise its image in $\text{Gal} (M /L)$ would fix a subfield of $M$ strictly larger than $L$ according to finite Galois theory, which would contradict our assumption that each element of $M \setminus L$ is moved by some element of $H$. In particular, some element of $H$ must map to the same element in $\text{Gal}(M/L)$ as $\sigma$. Hence $H$ contains an element of the coset $\sigma U_M$...
Question 1: Does "some element of $H$ must map to the same element in $\text{Gal}(M/L)$ as $\sigma$" mean there is some $\tau\in H$ such that $\tau|_{M}=\sigma$?
Question 2: Why does it imply $H$ contains an element of the coset $\sigma U_M$?
Answer 1: Correct, that's what it means.
Answer 2: The identity map $1_G$ of $K$ is an element of $U_M$. By the first answer $\tau\in\sigma U_M$.