I know that $\alpha \in \mathbb{F}_{p^n}$ is separable over $\mathbb{F}_p$ because finite fields are perfect. This means that the minimal polynomial $\mu_\alpha(x)\in\mathbb{F}_p[x]$ is separable. Is this true also on $\mathbb{F}_p(t)$ (viewing $\alpha$ as an element of $\mathbb{F}_{p^n}(t)$)?
My attempt was trying to see if $\mu_\alpha(x)$, visualized as an element of $\mathbb{F}_p(t)[x]$ is still an irreducible and separable polynomial, and it seems to me that I cannot use the extension $\mathbb{F}_p(t)/\mathbb{F}_p$ since is transcendental and not algebraic (so, a priori, it doesn't preserve the separability of elements).
Any ideas?
Yes. Any element $\alpha\in\Bbb{F}_{p^n}$ will also be separable over $\Bbb{F}_p(t)$. This follows from its separability over the prime field. Exactly due to the minimal polynomial $\mu_\alpha(x)$ over $\Bbb{F}_p$. When we replace $\Bbb{F}_p$ with any extension field, here $\Bbb{F}_p(t)$, it does not change the zeros of $\mu_\alpha(x)$ at all (how could it?). The polynomial $\mu_\alpha(x)$ may no longer be irreducible in general (here it does remain irreducible because we made a purely transcendental extension), but its zeros are still distinct. The new minimal polynomial is (in general) a factor of the old minimal polynomial, so its zeros continue to be distinct.
Generalization: If $\alpha$ is separable over a field $K$, and $L/K$ is a field extension, then $\alpha$ is separable over the field $L$ as well. The proof goes the same way. The minimal polynomial $m_L(x)$ of $\alpha$ over $L$ is a factor of the minimal polynomial $m_K(x)$ over $K$. The latter has no zeros with multiplicities $>1$, so the former cannot have either.