$\Omega$ is a set and $\mathcal{C}$ is a collection of subsets of $\Omega$. We call $\mathcal{C}$ is a semiring if and only if $\emptyset\in\mathcal{C}$ and $A,B\in\mathcal{C}\Rightarrow A\cap B\in\mathcal{C}, A\setminus B\in\mathcal{C}_{\Sigma f}$.
Now suppose that $\mathcal{C}$ is a semiring and $\mu$ is a finitely additive nonnegative set function on $\mathcal{C}$. Then we have the following conclusions:
(I) $\mu$ is monotone.
(II) If $A_n\in\mathcal{C},n\geq 1$, $A\in\mathcal{C}$ and $\sum_n A_n\subset A$, then $\sum_{n=1}^{\infty}\mu(A_n)\leq\mu(A)$.
The proof of (I) is as follows: Suppose $A,B\in\mathcal{C}$ and $A\subset B$, then $A\setminus B\in\mathcal{C}_{\Sigma f}$, i.e., there exists $A_i,1\leq i\leq m$ such that $B\setminus A=\sum_{i=1}^m A_i$. Therefore $$B=A\cup\sum_{i=1}^m A_i.$$ According to the finite additivity of $\mu$, we have $$\mu(B)=\mu(A)+\sum_{i=1}^m\mu(A_i)\Rightarrow\mu(A)\leq\mu(B),$$ which implies that $\mu$ is monotone.
But the proof of (II) is not easy. Your hints will be appreciated.
(I) As you noticed, one gets $\mu(B) \geqslant\mu(A)+\sum_{i=1}^m\mu(A_i)$ (whence $\mu(B) \geqslant \mu(A)$ indeed).
(II) Let $(A_n)$ be such that $\bigsqcup_n A_n \subset A$. The sequence $\Big(\sum_{n=1}^k \mu(An)\Big)_k$ is clearly increasing. Since it is bounded by $\mu(A)$ as well as increasing, it converges toward a limit $l \leqslant \mu(A)$.
Puting things together, we get:
$$\sum_{n=1}^\infty \mu(An) = \lim_{k \to \infty} \Big( \sum_{n=1}^k \mu(An) \Big) = l \leqslant \mu(A)$$