A result by Boston and Markin states that the minimum number of primes that ramify in a $G-$extension of $\mathbb{Q}$ is at least $d(G^{ab})$, the minimum number of generators of the abelianisation $G/[G,G]$.
In the case of $V_4$, which is abelian, and has a minimum of two generators, their result states that every $V_4-$ extension should be ramified for at least two primes.
I used SageMaths to calculate the discriminant of the number field $\mathbb{Q}(i+\sqrt{2})$ and verify that the Galois group is $V_4$. The discriminant is a power of 2, and so (2) is the only real prime that ramifies.
sage: R.<X> = PolynomialRing(QQ)
sage: K.<t> = NumberField(X^4-2*X^2+9)
sage: K.integral_basis()
[7/12*t^3 + 3/4*t^2 + 1/12*t + 1/4, 1/6*t^3 + 1/6*t, t^2, t^3]
sage: K.discriminant().factor()
2^8
sage: K.galois_group(type="pari")
Galois group PARI group [4, 1, 2, "E(4) = 2[x]2"] of degree 4 of the Number Field in t with defining polynomial X^4 - 2*X^2 + 9
How would I show that this extension also ramifies at the infinite prime?
It ramifies at the infinite prime because $\Bbb Q$ is totally real but $i+\sqrt 2$ is totally imaginary. (Recall that a real place $v$ in $K$ is said to ramify over $L$ if $v$ extends to a complex embedding that is not real.)