Polynomial close to separable polynomial in Gauss norm is separable itself.

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Let $(K,|\cdot |)$ be a complete non-archimedean valuation field. Let $f(x)\in K[x]$ be a separable polynomial of degree $n$. Let $||\cdot||$ be the Gauss norm associated with $|\cdot|$, i.e. $||f||=\max_{0\leq i\leq n}\{a_i\}$, where $a_i$ is the coefficient of degree $i$ term in $f$. I want to prove that there exists a $\delta>0$, such that for every polynomial $g(x)\in K[x]$ of degree $n$, $||f-g||<\delta$ implies $g$ is separable.

In the context the condition that $K$ is complete and non-archimedean is given, but I am also wondering are these two conditions necessary for this particular result?

Any help is much appreciated.

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Hint: Let's look at $n=2$ first. As we all learned in school (just did not phrase it that way), the polynomial $ax^2+bx+c$ is separable if and only if $b^2-4ac\neq0$. Now if that is the case, and we change $a,b,c$ "just a little", then $b^2-4ac$ will still be $\neq 0$.

Now that thing $b^2-4ac$ for quadratic polynomials is called the discriminant. Look up if such a thing exists for general polynomials, and what properties it has.