The first step to define cohomology of a profinite group $G$ is to consider discrete $G$-modules. These are abelian groups with the discrete topology and a continuous action $\psi \colon G \times M \rightarrow M$. It is a basic fact that $M$ being a discrete topological module should be equivalent to the fact that $\forall m \in M$ the stabilizer $\text{stab}(m)$ is an open subgroup of $G$ or equivalently $M=\bigcup_{U}M^U$ where $U$ ranges over all open subgroups of $G$ and $M^U=\{m \in M : u.m=m \}$.
I am ashamed to admit that I cannot see how the condition on open stabilizers implies $M$ must be discrete.
If $M$ is discrete then the preimage of $\{m\}$ along the restriction $G\times \{m\}\hookrightarrow G\times M \xrightarrow{\psi}M$ gives $\text{stab}(m)$ open. But I cannot see how to prove the other equivalence. Since this is a vary basic fact the proof should not involve more that basic point-set topology.
Edit: after consulting other notes I read that the condition $\text{stab}(m)$ open should be equivalent to the fact that the action map $\psi$ is continuous if we consider the module $M$ having the discrete topology. They indicate that the point is that openness of the stabilizers is equivalent to the continuity of the action map, not to the topology on $M$ being discrete.
This is very confusing: my first references were Serre's "Galois cohomology" and Symonds and Weigels "Cohomology of p-adic Analytic Groups" where the phrasing of the definition of discrete module seemed to indicate that the discreteness is equivalent to $\text{stab}(m)$ open for any $G$-module. Which is what I tried unsuccessfully to prove.
Can you confirm which version is the right approach?
Discrete $G$-modules have the discrete topology by definition, and then, as you say, openness of the stabilizers is equivalent to the continuity of the action map given this.
Discreteness is certainly not equivalent to the stabilizers being open. As a silly example, the trivial action of $G$ is continuous with any topology on $M$ whatsoever.