I think it is because $|1|_{2}=1$,
$|2|_{2}=2^{-1}\leq 1$ and
$|3|$,where 3 is prime
I am using the following Eisenstein criterion for $f(x)=a_{n}x^{n}+...+a_{0}$: $|a_{n}|=1$, $|a_{0}|=prime$ and $|a_{i}|<1$. On page 129 in http://www.jmilne.org/math/CourseNotes/ANT.pdf
As you can read from the comments, we have not agree if f is Eisenstein.
The original problem asks to prove $\mathbb{Q}_{2}[\sqrt{1+\sqrt{-2}}]$ is totally ramified. My approach was to show f is Eisenstein and then use the theorem below.
In the same page we have the following theorem: $K[\alpha]$ is totally ramified iff the minimal polynomial f of $\alpha$ over K is Eisenstein.
So $\mathbb{Q}_{2}[\sqrt{1+\sqrt{-2}}]$ minimal polynomial should be Eisenstein. Can you tell me what is going on?
Your polynomial $f(x)=x^4-2x^2+3$ is amenable to Eisenstein because $f(x+1)=x^4+4x^3+4x^2+ 2$, and this is an Eisenstein polynomial.