Let $\mathcal A$ be a collection of pairwise disjoint subsets of a $\sigma$-finite measure space,Show that $\mathcal A$ is at most countable.

Let $X$ be the space in question. Find a sequence of measurable sets $X_j$ of finite measure so that $X = \bigcup_j X_j$.

Suppose that $\mathcal{A}$ is uncountable. By the pigeon-hole principle, there is some $j$ so that uncountably many $A \in \mathcal{A}$ meets $X_j$ in a set of positive measure.

So, $\mu(X_j) \ge \sum_{A \in \mathcal{A}} \mu(X_j \cap A) = \infty$, which is a contradiction.

Note that the last fact follows from the observation that an uncountable sum of positive terms is infinite.

Edit in response to comments:

First the pigeon-hole principle. For each $A \in \mathcal{A}$ we have that $\mu(A) = \sum_j \mu(A \cap X_j)$. Since $\mu(A) > 0$, there is some $j_A$ so that $\mu(A \cap X_{j_A}) > 0$. Since there are an uncountable number of $A \in \mathcal{A}$ and only a countable number of $j$ in $\mathbb{N}$, there must be some $j \in \mathbb{N}$ so that $\mu(A \cap X_j) > 0$ for uncountably many $A \in \mathcal{A}$ (otherwise we'd have only countably many $A \in \mathcal{A}$...).

In response to the second comment, we use pairwise disjointness as follows:

$\mu(X_j) \ge \mu(\bigcup_{A \in \mathcal{S}} X_j \cap A) = \sum_{A \in \mathcal{S}} \mu(X_j \cap A)$, for any countable subset $\mathcal{S} \subset \mathcal{A}$.

Supping over all countable subsets $\mathcal{S} \subset \mathcal{A}$, we deduce the above.

Take $E_i$ a partition of the space with $\mu(E_i) < \infty$. How many elements of $\mathcal{A}$ can have measure $\mu(A \cap E_i)\geq 1/n$? Well, they are disjoint and $\mu(E_i)<\infty$, so am most finitely many, call them $\mathcal{A}_{i,n}$. Every $A\in \mathcal{A}$ has $\mu(A) > 0$, and hence $\mu(A \cap E_i) > 0$ for some $i$ and hence $\mu(A \cap E_i) \geq 1/n$ for $n$ large. It follows that $\mathcal{A} \subseteq \cup_{i,n} \mathcal{A}_{i,n}$ is contained (and hence equal to) a countable union of finite sets.