Polynomial with coefficients 'sufficiently close' to irreducible polynomial is irreducible In a paper I am reading, I think the following is implicitly used:

Let $(K, v)$ be a valued field, $f$ an irreducible, monic polynomial over $K$. If $g$ is another monic polynomial over $K$ such that the values of all coefficients of $f-g$ are 'sufficiently high', then $g$ must also be irreducible.

Is this true in general? How would one argue this? It feels intuitive as being irreducible is somehow an open property, but I do not see how to make this formal.
 A: This is based on a famous lemma for complete valued fields ( I forgot the author). Here is a  version of it.
Say we have $K \subset L$ a Galois extension of complete valued fields, $v$ in $L$ with conjugates $v=v_1$, $v_2$, $\ldots$, $v_m$ and $w\in L$ so that
$$|w - v| < |w - v_i|$$ for all $i\ne 1$. Then $v \in K(w)$. Indeed, consider an element $\sigma \in Gal(L/K)$ so that $\sigma(w) = w$. Note that $\sigma$ invariates the norm, since $K$ is complete. Then $\sigma(v)$ must be $v$. Indeed, it cannot be $v_i$, since $|w-v| < |w- v_i|$. 
Consequence: if $w \in K(v)$ is close enough to $v$ then $K(w) = K(v)$.
Let now $f$ the minimal polynomial of $f$. It has simple roots. By the implicit function theorem, if $g$ monic, with coefficients in $K$, is sufficiently close to $f$, the polynomial will have a root in $K(v)$ ( a complete field that contains the root $v$ of $f$).  Moreover, $w$ will be closer to $v$ than to the conjugates of $v$ ( in the normal closure of $K(v)$ ( for non-archimedean norm, the inequality $|w-v| < |v- v_i|$ for all $i \ne 1$ will do). We conclude that if $g$ sufficiently close to $f$, $g$ will have a root $w \in K(v)$, such that $K(w) = K(v)$. But then the degree of $w$ equals the degree of $v$. Therefore, $g$ has to be irreducible.    
