DavidL070949 said:

since the LC is technically not a tensor, do you still use the metric to raise and lower indices?

There is a Levi-Civita tensor ( https://en.wikipedia.org/wiki/Levi-Civita_symbol#Example:_Minkowski_space ),
which is the volume element ( https://en.wikipedia.org/wiki/Volume_element#Volume_element_of_manifolds ).
This is what is used in the definition of the magnetic field associated with a 4-velocity.

The signature-convention and the definitions of the electric and magnetic field given in the problem set agree with those in Wald's General Relativity (p. 64 Eq. 4.2.21 and 4.2.22, and Appendix B.2 for volume elements). (I didn't see these definitions of the fields in Carroll's text.)

I worked out the details, after some tedious calculations (and dummy-variable renaming).
I followed the approach by Geroch that I quoted in my Insight
https://www.physicsforums.com/insig...r-this-Spacetime-Approach-to-Electromagnetism ,
using Geroch's and Wald's conventions (the conventions of your problem)....
[which is different from the conventions used in my Insight (where I used (+---) instead of Wald's (-+++))].

I will give an outline (following Geroch's approach) here... (I can fill in the details later.)

Define ##h_{\alpha\beta} = g_{\alpha\beta} + U_\alpha U_\beta##, where ##h^\alpha{}_\beta## is a projection operator orthogonal to ##U^\alpha##.

Then, write
\begin{align}
F^{\alpha\beta} &= F^{\mu\nu} g^{\alpha}{}_{\mu} g^{\beta}{}_{\nu}\\
&= F^{\mu\nu} \left( h^{\alpha}{}_{\mu} - U^{\alpha} U_{\mu} \right) \left( h^{\beta}{}_{\nu} - U^{\beta} U_{\nu} \right)\\
& \qquad \mbox{
(use FOIL)}\\
&=
F^{\mu\nu} h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta
\end{align}

Over the years, it always bothered me that I got stuck here trying to express
##h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}## in terms of the epsilons so that I can use the definition of the magnetic field.
So, I looked online for some help.
From what I learned, I filled in the details that I needed to see.

I used their two-contraction identity ,
which I write for clarity as
##\epsilon^{\alpha\beta\rho\sigma} \epsilon_{\mu\nu\rho\sigma}
= (-1)^s\ 2 \left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} \right)
\stackrel{s=1}{=} 2 \left( \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} - \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} \right)##, where ##s## is the number of minus-signs in the metric).
It was helpful to write this in alternative notations
\begin{align}
\epsilon^{\alpha\beta\rho\sigma} \epsilon_{\mu\nu\rho\sigma}
= (-1)^s\ 2 \left( \delta^{\alpha}_{\mu} \delta^{\beta}_{\nu} - \delta^{\alpha}_{\nu} \delta^{\beta}_{\mu} \right)
= (-1)^s\ 2\ \delta^{\alpha\beta}_{\mu\nu}
= (-1)^s\ 2
\begin{vmatrix}
\delta^\alpha_\mu & \delta^\alpha_\nu \\
\delta^\beta_\mu & \delta^\beta_\nu
\end{vmatrix}
\end{align}

I also found it helpful to use a one-contraction identity, which I write as
\begin{align}
\def\ALPHA{{\color{red}\alpha}}
\def\BETA{{\color{red}\beta}}
\def\MU{{\color{red}\mu}}
\def\NU{{\color{red}\nu}}
\epsilon^{\alpha\beta\gamma\sigma} \epsilon_{\mu\nu\rho\sigma}=(-1)^s\ \delta^{\alpha\beta\gamma}_{\mu\nu\rho}
= (-1)^s
\begin{vmatrix}
\delta^\alpha_\mu & \delta^\alpha_\nu & \delta^\alpha_\rho \\
\delta^\beta_\mu & \delta^\beta_\nu & \delta^\beta_\rho \\
\delta^\gamma_\mu & \delta^\gamma_\nu & \delta^\gamma_\rho
\end{vmatrix}
\end{align}

Here is the first connection to relate the h to the epsilon.
Use the two-contraction identity to show
\begin{align}
\epsilon^{\ALPHA\beta\rho\sigma} \epsilon_{\MU\nu\rho\sigma} U_\beta U^\nu
&=
2 \left( h^{\ALPHA}{}_{\MU} \right)
\end{align}

In forming ##h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}##, one gets
\begin{align}
h^{[\ALPHA}{}_{[\MU} h^{\BETA]}{}_{\NU]}
&=
\delta^{[\ALPHA|}_{[\MU|} \delta^{|\BETA]}_{|\NU]}
+\delta^{[\ALPHA|}_{[\MU|} U^{|\BETA]} U_{|\NU]}
+ U^{[\ALPHA|} U_{[\MU|} \delta^{|\BETA]}_{|\NU]}
\end{align}.
Then "a miracle occurs".
This (from my online resource) turns out to be equal to ##\frac{1}{2}\epsilon^{\ALPHA\BETA\gamma\omega}\epsilon_{\MU\NU\pi\omega} U_\gamma U^\pi##,
shown by using the one-contraction formula.
(It works, but I'm not a fan of this approach.
I prefer some more motivation from the antisymmetrized-hh (using the two-contraction identity)
to a more-recognizable expression using the one-contraction identity.
I wonder if this can be written as a Laplace-expansion of a determinant involving the deltas.)

With this, I finally get
\begin{align}
F^{\alpha\beta} &= F^{\mu\nu} g^{\alpha}{}_{\mu} g^{\beta}{}_{\nu}\\
&=
F^{\mu\nu} h^{[\alpha}{}_{[\mu} h^{\beta]}{}_{\nu]}
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta \\
&=
F^{\mu\nu} \left( \frac{1}{2}\ \epsilon^{\ALPHA\BETA\gamma\omega}\epsilon_{\MU\NU\pi\omega} U_\gamma U^\pi \right)
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta \\
&=
\epsilon^{\ALPHA\BETA}{}_{\gamma\omega} U^\gamma \left( B^\omega \right)
+ U^{\alpha}E^{\beta}
- E^{\alpha}U^\beta
\end{align}