The multiplication-by-$m$ map $[m]:E\to E$ is unramified, so there exists a finite set of primes $S$, depending only on $E$ and $m$, so that for every $P\in E(K)$, the field generated by the coordinates of the points in $[m]^{-1}(P)$ is unramified outside of $S$. (This is where we use Chevelley-Weil.) The degree of that extension is also bounded as a function of $m$. It follows from standard results in algebraic number theory that there are only finitely many such fields. Hence $[m]^{-1}\bigl(E(K)\bigr)$ is contained in $E(L)$, where the extension $L/K$ is a **finite** extension.

I'll let you read elsewhere the fact that it suffices to prove the weak Mordell-Weil theorem under the assumption that $E[m]\subset E(K)$. Under that assumption, and with notation as in the first paragraph with $L=K\bigl([m]^{-1}\bigl(E(K)\bigr)\bigr)$, there is a well-defined injective homomorphism
$$
E(K)/mE(K) \longrightarrow \operatorname{Hom}\bigl(\operatorname{Gal}(L/K),E[m]\bigr)
$$
defined by
$$ P \longmapsto \bigl(\sigma\mapsto \sigma(Q)-Q\bigr)
\quad\text{for any choice of $Q\in E(L)$ satisfying $[m](Q)=P$.}
$$
Since $L/K$ is a finite extension from Chevalley-Weil, and since $E[m]$ is a finite group, this gives the finiteness of $E(K)/mE(K)$.