How to apply Hilbert's Irreducibility theorem in this proof? I'm going over a paper and I've come over this particular proof:

We need the following result which is probably well known (however, I could not find a good reference).
Lemma. Let $E$ be a number field and let $X$ be a smooth projective geometrically connected curve over $E$. Let $\pi:X'\to X$ be a connected Galois covering of degree $d$ which splits completely (i.e. every closed point $P\in X$ has exactly $d$ pre-images in $X'$). Then, $d=1$.
Proof. Let $f:X\to\mathbf{P}^1_E$ be a non-constant morphism and let $f'=\pi\circ f:X'\to\mathbf{P}_E^1$. By Hilbert's Irreducibility Theorem there exists a closed $E$-rational point $P$ of $\mathbf{P}_E^1$ which has only one pre-image under $f'$. Therefore it has only one pre-image under $f$, say $P'$, and $P'$ has only one pre-image under $\pi$. Hence $d=1$.

First of all, I didn't know what a Galois covering exactly is. Turns out, apparently it's just a finite surjective morphism $X\to Y$, such that the induced function field extension $K(Y)/K(X)$ is Galois.
So, the question is, how exactly can I apply Hilbert's Irreducibility Theorem in this case? The connection between its rather elementary formulation and this algebraic geometry context eludes me. Is there maybe a generalization or an equivalent formulation in the language of algebraic geometry?
The conclusion of the argument is then clear.
 A: The function field of $\mathbf P^1_E$ is $E(T)$, where $T$ is an indeterminate. Let $K$ be the function field of the algebraic curve $X$. A non-constant map $$\varphi: X \rightarrow  \mathbf P^1_E$$ induces an embedding $$ E(T) \rightarrow K$$ and, since $X$ is a curve over $E$, the extension $K/E(T)$ is finite. Moreover, since $E$ has characteristic zero, $K/E(T)$ is also separable and, by the primitive root theorem, there exists $a \in K$ such that $K = E(T)(a)$. Let $f$ be the minimal polynomial of $a$ over $E(T)$.
By Hilbert's irreducibility theorem, there exists $e \in E$ such that the specialization $f(e, A)$ is irreducible. Let $P$ be the $E$-rational point of $\mathbf P^1_E$ corresponding to $e$. A point in the fiber $\varphi^{-1}(P)$ thus corresponds to a place of $K$ above the place $T - e$ of $E(T)$.

Claim: The place $T - e$ of $E(T)$ is inert in the extension $K/E(T)$. In particular, $ \varphi^{-1}(P)$ consists of a single point.

Proof: Since the class of $a$ in $E[T, a]/(T - e)$ has minimal polynomial $f(e, A)$ over the field $E[T]/(T - e)$, the quotient $E[T, a]/(T - e)$ is a field extension of  $E[T]/(T - e)$ of degree $[K:E(t)]$. Therefore, $T - e$ is inert in $K/E(T)$.
