Let $M/K$ be arbitrary extension. Then, we can define Krull topology on $Gal(M/K)$ by taking {$Gal(M/F)|K⊆F⊆M,[F:K]＜∞$} as fundamental system of neighborhood of $0$.
I guess $Gal(M/K)$ has discrete topology if only if $M/K$ is finite extension, because krull topology is one which is also defined as inverse limit of discrete topological group.
But I want to prove from definition of krull topology.
The condition of krull topology is discrete is equivalent to $∀σ∈Gal(M/K)$,{$σ$} is open subset.This is equivalent to $∀σ∈Gal(M/K)$,$∃V∈${$σGal(M/K)|K⊆F⊆M$}$,V⊆Gal(M/K)$ from definition of fundamental system of neighborhoods. But I cannot proceed from here. Why this is equivalent to $M/K$ is finite ?
IMO, you really should only be talking about "Galois group" and "Krull topology" when your extension is actually Galois, because the resulting topological group can be very badly behaved otherwise.
For example, if $M/K$ is purely inseparable of infinite degree (e.g. $\mathbb{F}_p(X_n;n\in\mathbb{N})/\mathbb{F}_p(X_n^p;n\in\mathbb{N})$), then $\mathrm{Aut}(M/K)$ is trivial, so necessarily discrete, yet not finite by assumption. Thus, your claim is not true in general.
Now, assume $M/K$ is a Galois extension. Since $\mathrm{Gal}(M/K)$ is a topological group (homogeneous suffices), it is discrete if and only if $\{e\}\subseteq\mathrm{Gal}(M/K)$ ($e$ is the identity element) is open. By the definition via the fundamental system of neighborhoods, if this is the case, there is an intermediate extension $M/F/K$ finite over $K$, such that $\mathrm{Gal}(M/F)\subseteq\{e\}$. Since $e\in\mathrm{Gal}(M/F)$, it follows that $\mathrm{Gal}(M/F)=\{e\}$. Since $M^{\{e\}}=M$ trivially and $M^{\mathrm{Gal}(M/F)}=F$ by Galois theory, it follows that $M=F$ is finite over $K$. The converse is hopefully clear.