If $A$ is a Dedekind domain and $I \subset A$ a non-zero ideal, then every ideal of $A/I$ is principal.

Yes. Factor $I = \displaystyle\prod_{i =1}^n \mathfrak{p}_i^{e^i}.$ Then by the Chinese Remainder Theorem $A/I \cong \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}.$ So it is enough to show each factor $A/\mathfrak{p}_i^{e^i}$ is principal. The ideals of $A/\mathfrak{p}_i^{e^i}$ are exactly the images of the ideals of $A$ containing $\mathfrak{p}_i^{e^i},$ i.e., $\mathfrak{p}_i^{n}$ for $1\le n \leq e_i,$ under the projection map $\pi:A \rightarrow A/\mathfrak{p}_i^{e^i}.$ If $\pi(\mathfrak{p}_i) = \pi(\mathfrak{p}_i^2)$ then $\pi( \mathfrak{p}_i) = 0$ and $A/\mathfrak{p}_i^{e_i}$ is a field. Otherwise, let $\alpha \in\pi( \mathfrak{p}_i) \setminus \pi(\mathfrak{p}_i^2).$ Then $(\alpha)$ is a proper ideal such that $(\alpha) \not\subset \pi(\mathfrak{p}_i^n) $ for any $n\ge 2.$ It follows $(\alpha) = \pi(\mathfrak{p}_i).$ We conclude $\pi(\mathfrak{p}_i^n) = (\alpha^n)$ and hence $A/\mathfrak{p}_i^{e_i}$ is principal.

HINT $\:$ By the approximation theorem, Dedekind domain ideals $\rm\:J\!\ne\! 0\:$ are strongly two generated, i.e. for each $\rm\:0\ne i\in J\:$ there exists $\rm\:j\in J\:$ such that $\rm\:J = (i,j)\:.\:$ Therefore every ideal $\rm\: J\supset I\:$ can be presented in the form $\rm\:J = (i,j)\:$ for $\rm\:0\ne i\in I,\ j\in J\:.\:$ Hence $\rm\ J = (j)\pmod I\ $ is principal.

NOTE $\ $ This property characterizes Dedekind domains, i.e. a domain $\rm\:D\:$ is Dedekind iff every nonzero ideal of $\rm\:D\:$ is strongly two generated.