Is an exact functor to FSet faithful?

No, a left exact functor to the category of finite sets need not be faithful. As a trivial example, the functor $\mathcal{C} \to \mathrm{FinSet}$ sending everything to a singleton is left exact, but not faithful unless $\mathcal{C}$ is equivalent to a meet-semilattice. For a less trivial example, the functor $\mathrm{FinCat} \to \mathrm{FinSet}$ that sends a finite category to its set of objects is also left exact, but not faithful.

Of course, neither of the above examples is conservative either. An example of a faithful non-conservative left exact functor to the category of finite sets is the forgetful functor from the category of finite topological spaces.

Yes, conservativity is needed for this result. The problem with your original argument is that it tries to conclude from the fact that $F(\mathrm{id}_X)$ is an equalizer of $F(\phi)$ and $F(\psi)$ that $\mathrm{id}_X$ is an equalizer of $\phi$ and $\psi$. But this reasoning is backwards: you are entitled to use the converse reasoning, so that if $f:E\to X$ is an equalizer of $\phi$ and $\psi$ then $F(f)$ is an equalizer of $F(\phi)$ and $F(\psi)$. So this shows that $F(f)$ is an isomorphism, where your argument about conservativity comes in.

Note that no use is made of a particular codomain here: any conservative functor between categories with equalizers which preserves equalizers is faithful.