I am trying to understand the proof in Bushnell and Henniart - The local Langlands conjecture for $\operatorname{GL}_2$ of the fact (§28.6, Lemma 1) that smooth irreducible representations of the Weil group are finite dimensional. I have several doubts about the proof. The proof is:

Let $(\rho,V)$ be a smooth irreducible representation of the Weil group $W_F$ of a non Archimedean local field $F$. Let $v\in V$, $v \neq 0$. $V$ is irreducible, so the elements $\rho(x)v$ with $x \in W_F$ generate all $V$. Since $V$ is smooth, $v$ is fixed by an open subgroup of $W_F$ and so by an open subgroup of the inertia group $I_F$. These groups are of the form $J=I_F \cap \Omega_E$ where $\Omega_E$ is an algebraic closure of a finite extension $E$ of $F$. WLOG we can suppose $E/F$ Galois and so $J$ normal in $I_F$ and so in $W_F$. Since it is normal in $W_F$, then it is in the kernel of the representation. Now, let $\phi$ be a Frobenius element in $W_F$. It acts by conjugation on the finite group $I_F/J$, so some power of $\phi^d$ acts trivially. Thus $\rho(\phi^d)$ commutes with $\rho(W_F)$ and so, by Schur, $\rho(\phi^d)$ must be scalar. Hence, the vectorial space generated by $\rho(x)$ is finite dimensional and so is $V$.

My doubts are

- Why being normal in $W_F$ implies being in the kernel?
- I am not totally sure about what we are doing with the Frobenius element to say that $\rho(\phi^d)$ acts trivially and hence we can use Schur.
- The conclusion: how is it enough?