Since $\mu$ is a finitely–additive measure, to prove countable additivity it will suffice to show that if $A_1, A_2, ...$ are a decreasing sequence of sets in $F$, where $F$ is the $\sigma$-algebra with $\inf_n \mu(A_n) > 0$ then $\cap_{n \geq 1} (A_n) \neq \phi$.
Require hints why it suffices.
But we know the fact:
Suppose $P$ is a finitely additive set function on a $\sigma$-algebra $\mathscr A$ such that $P(\Omega) = 1, P(A) \geq 0$ for every $A \in \mathscr A$ and is continuous; that is for any decreasing sequence $(B_n), B_n \in \mathscr A$, if $B = \cap_{n=1}^{\infty}B_n$, we have $P(B) = \lim_{n\to \infty} B_n$. Then $P$ is countably additive.
Can we relate the two fact?
Suppose $\{E_n\}$ is a disjoint sequence of measurable sets with union E and $\mu (E) \neq \sum_1 ^{\infty} \mu (E_n)$. Let $A_n =E - \{E_1 \cup E_2 \cup ... \cup E_n\}$. Then $A_n$ is decreasing and their intersection if empty. Now $E$ is contained in the union of $A_n$ and $ \{E_1 \cup E_2 \cup ... \cup E_n\}$ so $\mu (E) \leq \mu (A_n) + \sum_1 ^{n} \mu (E_i) < \sum_1 ^{\infty} \mu (E_i) $. Hence $\mu (E) \leq inf(\mu (A_n)) + \sum_1 ^{\infty} \mu (E_i)$. We must have $\mu (E) < \sum_1 ^{\infty} \mu (E_i) $ because $\sum_1 ^{k} \mu (E_i) \leq \mu (E)$ for each k which shows $\sum_1 ^{\infty} \mu (E_i) \leq \mu (E)$. W ehave now proved that $inf \mu (A_n) >0$. though intersection of the $A_n$'s is empty, contradicting the hypothesis.