Prove roots of $x^5 - 1$ are constructible

Since $[F : \mathbb Q] = 4$, it suffices to find an intermediate extension $ \mathbb Q \subset K \subset F$.

Then $[K : \mathbb Q] = 2$ and $[F : K] = 2$ implies that they are constructible.

As mentioned in the comments, $K=\mathbb Q(z+\bar z)$ is a natural candidate.

Indeed, let $w = z+\bar z$. Since $\bar z=z^4$, we have $wz=z^2+1$ and thus this is the minimal equation for $z$ over $K=\mathbb Q(w)$.

Hint

Since the equation $x^4+x^3+x^2+x+1=0$ is symmetric, the standard substitution $t=x+\frac{1}{x}$ solves it. Note that at this point you already know that $x$ is constructible: the substitution leads to an equation of degree $\frac{4}{2}=2$ isn $t$ with rational coefficients, and hence $t$ is constructible. Moreover, $x$ is the root of the quadratic $x^2-tx+1=0 \in \mathbb Q(t)[X]$ and thus constructible.

But, just to see here are the details $$t=x+\frac{1}{x}\\ t^2-2=x^2+\frac{1}{x^2}$$

Then $$x^4+x^3+x^2+x+1=0 \Rightarrow x^2+x+1+\frac{1}{x} +\frac{1}{x^2}=0 \Rightarrow x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0 \Rightarrow \\ t^2-2+t+1=0 \Rightarrow t^2+t-1=0$$

Just solve for $t$, and then solve $$t=x+\frac{1}{x}\Rightarrow x^2-tx+1=0$$