That is not true. Let $A$ be a polynomial ring in two variables, $k[s,t]$. Let $C$ be the $A$-subalgebra of the fraction field generated by the fraction $s/t$, i.e., $C=k[(s/t),t]$. Both $A$ and $C$ are unique factorization domains (roughly by Gauss's Lemma). Since $A$ is integrally closed in its fraction field, the extension $A\to C$ is "root closed".
Let $w$ be the element $(s/t)^2-t$, just for one example. Then the $A$-algebra $A[w]$ equals $k[s,t][w]/\langle t^2w + (t^3-s^2) \rangle$. The quotient of $A[w]$ by the principal ideal generated by $w$ equals $k[s,t]/\langle t^3-s^2 \rangle$. This is an integral domain. Thus, the principal ideal generated by $w$ in $A[w]$ is a prime ideal. Of course $w$ is prime, hence irreducible, in $C$ since $C/\langle w\rangle$ is the polynomial ring $k[s/t]$.
Finally, $A[w]$ is not a unique factorization domain. In fact, it is not even integrally closed in its fraction field. The element $f=s/t$ of the fraction field satisfies the monic polynomial, $$f^2 -(w+t) = 0.$$ If $f$ were an element of $A[w]$, then $A[w]$ would equal all of $C$. However, $A[w]/\langle w \rangle$ equals $k[s,t]/\langle t^3-s^2\rangle$. This is not regular, it is not a unique factorization domain, it is not integrally closed in its fraction field, etc. On the other hand, $C/\langle w \rangle$ is isomorphic to $k[s/t]$. This is regular, it is a unique factorization domain, it is integrally closed in its fraction field, etc.