Like you said "$A_\mu$ some dynamical $U(1)$ gauge field that minimally couples to $\phi$". It means that the covariant derivative is : $$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$ with $q$ the $U(1)$ charge of the scalar field. As a consequence, if $\phi$ is not $U(1)$ charged you will not have the second term in the covariant derivative and hence $D_\mu$ is equivalent to the standard derivative $\partial_\mu$. When the scalar field $\phi$ is VEVed with fluctuations around this VEV : $$\phi = \frac{(v+h_1)+ih_2}{\sqrt{2}}$$ and that you compute the kinetic term $(D_\mu \phi)^\dagger D_\mu \phi$, you will get a mass term for $A_\mu$ which is : $$\frac{1}{2}q^2v^2A_\mu A^\mu$$ which is proportional to $q$ the $U(1)$ charge of $\phi$ and this term appears in the contraction of the second terms of $(D_\mu \phi)^\dagger$ and $D_\mu \phi$. If these second terms was not here, i.e. if the scalar field $\phi$ was not $U(1)$ charged, then the gauge field $A_\mu$ can't get a mass.

Finally, the requirement that $\phi$ is $U(1)$ charged enters when you require a non-trivial minimal coupling between $A_\mu$ and $\phi$.

$\textbf{EDIT}$ :

If you've already seen the Glashow-Weinberg-Salam (GWS) Model, the analogy is that the Higgs field is $SU(2)$ charged (because it's an $SU(2)$ doublet). Thus, when the Higgs field is VEVed, the gauge fields acquire a mass.

Like you said "$A_\mu$ some dynamical $U(1)$ gauge field that minimally couples to $\phi$". It means that the covariant derivative is : $$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$ with $q$ the $U(1)$ charge of the scalar field. As a consequence, if $\phi$ is not $U(1)$ charged you will not have the second term in the covariant derivative and hence $D_\mu$ is equivalent to the standard derivative $\partial_\mu$. When the scalar field $\phi$ is VEVed with fluctuations around this VEV : $$\phi = \frac{(v+h_1)+ih_2}{\sqrt{2}}$$ and that you compute the kinetic term $(D_\mu \phi)^\dagger D_\mu \phi$, you will get a mass term for $A_\mu$ which is : $$\frac{1}{2}q^2v^2A_\mu A^\mu$$ which is proportional to $q$ the $U(1)$ charge of $\phi$ and this term appears in the contraction of the second terms of $(D_\mu \phi)^\dagger$ and $D_\mu \phi$. If these second terms was not here, i.e. if the scalar field $\phi$ was not $U(1)$ charged, then the gauge field $A_\mu$ can't get a mass.

Finally, the requirement that $\phi$ is $U(1)$ charged enters when you require a non-trivial minimal coupling between $A_\mu$ and $\phi$.

$\textbf{EDIT}$ :

If you've already seen the Glashow-Weinberg-Salam (GWS) Model, the analogy is that the Higgs field is $SU(2)_L\times U(1)_Y$ charged (because it's an $SU(2)$ doublet and it has an hypercharge $Y$). Thus, when the Higgs field is VEVed, the gauge fields acquire a mass.

Like you said "$A_\mu$ some dynamical $U(1)$ gauge field that minimally couples to $\phi$". It means that the covariant derivative is : $$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$ with $q$ the $U(1)$ charge of the scalar field. As a consequence, if $\phi$ is not $U(1)$ charged you will not have the second term in the covariant derivative and hence $D_\mu$ is equivalent to the standard derivative $\partial_\mu$. When the scalar field $\phi$ is VEVed with fluctuations around this VEV : $$\phi = \frac{(v+h_1)+ih_2}{\sqrt{2}}$$ and that you compute the kinetic term $(D_\mu \phi)^\dagger D_\mu \phi$, you will get a mass term for $A_\mu$ which is : $$\frac{1}{2}q^2v^2A_\mu A^\mu$$ which is proportional to $q$ the $U(1)$ charge of $\phi$ and this term appears in the contraction of the second terms of $(D_\mu \phi)^\dagger$ and $D_\mu \phi$. If these second terms was not here, i.e. if the scalar field $\phi$ was not $U(1)$ charged, then the gauge field $A_\mu$ can't get a mass.

Finally, the requirement that $\phi$ is $U(1)$ charged enters when you require a non-trivial minimal coupling between $A_\mu$ and $\phi$.

Like you said "$A_\mu$ some dynamical $U(1)$ gauge field that minimally couples to $\phi$". It means that the covariant derivative is : $$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$ with $q$ the $U(1)$ charge of the scalar field. As a consequence, if $\phi$ is not $U(1)$ charged you will not have the second term in the covariant derivative and hence $D_\mu$ is equivalent to the standard derivative $\partial_\mu$. When the scalar field $\phi$ is VEVed with fluctuations around this VEV : $$\phi = \frac{(v+h_1)+ih_2}{\sqrt{2}}$$ and that you compute the kinetic term $(D_\mu \phi)^\dagger D_\mu \phi$, you will get a mass term for $A_\mu$ which is : $$\frac{1}{2}q^2v^2A_\mu A^\mu$$ which is proportional to $q$ the $U(1)$ charge of $\phi$ and this term appears in the contraction of the second terms of $(D_\mu \phi)^\dagger$ and $D_\mu \phi$. If these second terms was not here, i.e. if the scalar field $\phi$ was not $U(1)$ charged, then the gauge field $A_\mu$ can't get a mass.

Finally, the requirement that $\phi$ is $U(1)$ charged enters when you require a non-trivial minimal coupling between $A_\mu$ and $\phi$.

$\textbf{EDIT}$ :

If you've already seen the Glashow-Weinberg-Salam (GWS) Model, the analogy is that the Higgs field is $SU(2)$ charged (because it's an $SU(2)$ doublet). Thus, when the Higgs field is VEVed, the gauge fields acquire a mass.