Nedoma's pathology says that if there is a measurable space $(X,F)$ with $|X|>2^{\aleph_0}$ then the diagonal is not measurable in $X\times X$. This proof relies heavily on the fact that we require sigma algebras to be countably additive.
We don't want sigma algebras to be arbitrarily additive because if points are measurable, then everything is measurable.
The main measure space we care about is $\Bbb R$, so we require countable additivity as countability is cardinality lower than the cardinality of the reals.
If we change the definition of sigma algebra to:
Let $X$ be a set with infinite cardinality $|X|$, a $\sigma$ algebra on $X$ is a subset $\sigma \subseteq \mathcal P(X)$ satisfying:
$\emptyset, X\in \sigma$
If $A\in \sigma$ then $X\setminus A\in \sigma$.
If $\{A_\alpha\}$ is a collection of sets with $|\{A_\alpha\}|<|X|$ then $\cup A_\alpha\in \sigma$
do we avoid pathologies like Nedoma's pathology? Does this introduce any new pathologies?
This causes far more problems than it solves. For one thing, the entire theory of (say) Lebesgue measure would fall apart unless you make some major set-theoretic assumptions, since you can't prove in ZFC that the algebra of Lebesgue measurable sets is a Lebesgue $\sigma$-algebra by your definition.
Even assuming (say) CH, consider what happens if you try to take a product measure space $[0,1]^I$ of copies of Lebesgue measure where the index set $I$ has cardinality at least $2^{\aleph_0}$. Then by definition, the "product $\sigma$-algebra" would have to be closed under $2^{\aleph_0}$-sized unions, since $|[0,1]^I|>2^{\aleph_0}$. But that means that for any $i\in I$ and any $A\subseteq[0,1]$, the set $A\times[0,1]^{I\setminus\{i\}}$ would have to be measurable, since it is the union of the measurable sets $\{a\}\times [0,1]^{I\setminus\{i\}}$ for each $a\in A$. Since the measure of such a set ought to be the same as the measure of $A$ in $[0,1]$, this means that there would be no reasonable way to define a product measure on the entire "product $\sigma$-algebra", since not every subset of $[0,1]$ is measurable. Also, for instance, this would mean that the diagonal map $[0,1]\to [0,1]^I$ is not measurable.