This is Exercise 20.2 in Matsumura's Commutative Ring Theory:
Let $A$ be an integral domain. We say that $A$ is locally UFD if $A_{\mathfrak m}$ is a UFD for every maximal ideal $\mathfrak m$. If $A$ is a semilocal integral domain and $A$ is locally UFD, then $A$ is a UFD.
Here is my idea, but I realized there is a gap. I want to show that $A$ satisfies the ascending chain condition on principal ideals, and every irreducible element is prime (this is exercise 1.4 in Matsumura). The ascending chain condition is easy: If one has a chain $(a_1) \subset (a_2) \subset \dotsc$, it stabilises after localisation wrt any maximal ideal $\mathfrak m$ (because $A_{\mathfrak m}$ is a UFD), say after $n_{\mathfrak m}$ steps. Since there are only finitely many maximal ideals, there is an upper bound $n$ for this. Then for $k \geq n$, $$(a_{k+1}) / (a_k)$$ is zero after localisation wrt any maximal ideal, hence is zero as an $A$-module. So we get the ascending chain condition on principal ideals.
$\newcommand\divides{\,\vert\,}$ Now suppose $a \in A$ is an irreducible element, and $a \divides p \cdot q$.
Question / Gap: Is it true that $\frac a 1$ is irreducible in $A_{\mathfrak m}$, provided that $a \in \mathfrak m$?
Suppose this is true. Then we know that $\frac a 1$ is prime, so wlog we may assume $\frac a 1 \divides \frac p 1$, say $$\frac a 1 \cdot \frac r s = \frac p 1.$$ So $ar = ps$ for some $s \notin \mathfrak m$. Let $\mathfrak m_1, \dotsc, \mathfrak m_n$ be the maximal ideals of $A$. By the above, we get elements $r_i \in A, s_i \notin \mathfrak m_i$ with $$a r_i = p s_i.$$ (If $a \notin \mathfrak m$, simply take $r_i = p, s_i = a$). In particular, the ideal $(s_1, \dotsc, s_n)$ is not contained in any maximal ideal, so $$(s_1, \dotsc, s_n) = A.$$ We can write $1 = \sum_i c_i s_i$ for some $c_i \in A$, and hence $$p = p \sum_i c_i s_i = a \sum_i c_i r_i,$$ so $a \divides p$.
Is it possible to fill the gap above? I know that in general irreducible elements may become reducible after localisation...
If not, how else could one tackle this problem?