If you have any sigma-algebra $A$ on some finite set $X$ you can partition $X$ and use the disjoint subsets of $X$ to generate $A$. Why cannot you do the same if $X$ was infinite?
I was thinking, it's because you would need uncountably many disjoint subsets. But isn't that how the Borel algebra on $\mathbb{R}$ is generated? There are uncountably many open intervals on $\mathbb{R}$.
So what is so different about other sigma algebras on infinite sets?
On
You can use disjoint subsets as generators in certain cases, for example if $X$ is countable, or depending on the structure of the $\sigma-$algebra. In general though the problem is that sigma algebras are defined only through countable operations on the generators. Here is an example which might address your question more precisely.
Take the real line $\mathbb{R}.$ Then suppose you take a set of generators $\{ \mathcal{A}_i \}_{i \in I}$ that are disjoint from each other. Note that we would like the Borel sigma algebra to contain singletons, and note that if $$ \mathcal{A} \in \sigma (\mathcal{A_i})$$ with $\mathcal{A} \subset \mathcal{A}_{i_0}$ then by the disjointness either $\mathcal{A} = \mathcal{A_i}$ or $\mathcal{A} = \emptyset.$ So for the singletons to be in the generated sigma algebra, they need to be themselves the generators. But the sigma algebra generated by the singletons is strictly smaller than the one generated by the open intervals. See for example this question.
Note on the other hand, that this does not mean that the generators should be uncountable. Indeed the set of all open intervals with rational radius and rational center form a set of generators for the Borel sigma algebra.
It's not all infinite $X$ it doesn't work for. For example, every $\sigma$-algebra on $\mathbb Q$ is generated by a partition -- and likewise for any countable $X$.
The underlying reason is that a $\sigma$-algebra is only supposed to be closed under countable unions or intersections.
This has two effects. First, the argument that produces a partition form an arbitrary $\sigma$-algebra depends on taking intersections that may have as many operands as there are elements in $X$. When we try to apply it to an algebra on $\mathbb R$, we do get disjoint subsets out of it, but those disjoint subsets are not necessarily in the $\sigma$-algebra we started with.
Second, even if the partitions we get are in the $\sigma$-algebra (as is the case for the Borel algebra on $\mathbb R$), they may not generate it. In the case of the Borel algebra, the partitions are just the singletons, but a random Borel set cannot be generated as a countable union of singletons.
(By the way, the Borel algebra doesn't need uncountably many generators -- it is enough to take the open intervals with rational endpoints).