This is theorem 10.1 of https://math.mit.edu/classes/18.785/2015fa/LectureNotes10.pdf
Let $A$ be a complete DVR with fraction field $K$ (assume $K$ is perfect if necessary) and residue field $k$. Let $l/k$ be a finite separable extension. By the primitive element theorem, write $l = k[x] / g_0$ for some $g_0 \in k[x]$ irreducible and monic. Lift $g_0$ to some $g \in A[x]$, also monic and irreducible (as a factorization of $g$ would correspond to a factorization of $g_0$).
Now, let $L = K[x] / g$, which is also a field as $g$ is monic and irreducible. Then, $L/K$ is also separable. Let $B$ be the integral closure of $A$ in $L$ (by a previous theorem, Theorem 9.25, $B$ is the valuation ring of $L$).
Let $a = x + (g) \in L$. Then, the claim is that $B = A[a]$. The notes just says that this is "by construction, it is not something we need to prove. ".
I don't understand why this is true. We need to use that $K$ is the field of fractions of a complete DVR, because this is false if that hypothesis is dropped. For example, $K = \mathbb{Q}$ and $a = \sqrt{5}$. Why does $B = A[a]$?