Galois Basic Question (Cyclic Order 4 Extension of Q cannot contain i)

(fleshing this out now that OP worked out the details themself)

Because $K/\Bbb{Q}$ is Galois and complex conjugates share a minimal polynomial over $\Bbb{Q}$, we see that the complex conjugate of any element of $K$ is also an element of $K$. Therefore complex conjugation, call it $\sigma$, is an element of the Galois group $G=\operatorname{Gal}(K/\Bbb{Q})$.

Assume contrariwise that $i\in K$. Because $\sigma(i)=-i$ and $\sigma^2=1_G$, we can conclude that $\sigma$ is of order two. As $G$ is assumed to be cyclic, $H=\langle\sigma\rangle$ is the only subgroup of index two in $G$. By Galois correspondence the fixed field $M:=K^H=K\cap\Bbb{R}$ is the only quadratic intermediate field $\Bbb{Q}\subset M\subset K$.

But $\Bbb{Q}(i)$ would also be a quadratic intermediate field contradicting the above. Hence $i\notin K$.

This argument is not correct. All you've shown is that the restriction of $\sigma$ to $\mathbb{Q}(i)$ cannot have order $4$. But the case that $\sigma(i)=-i$ is (a priori) possible even if $\sigma$ has order $4$: in this case, you just would conclude that $\mathbb{Q}(i)$ is the fixed field of $\sigma^2$. This doesn't (obviously) mean that $\sigma^2$ is the identity on all of $K$.

Here is a sketch of a correct argument. We must have $K=\mathbb{Q}(i,\sqrt{a})$ for some $a\in\mathbb{Q}(i)$ that is not a square. In order for this $K$ to be Galois over $\mathbb{Q}$, $\sqrt{\overline{a}}$ must also be in $K$. Writing down explicitly what an element of $K$ looks like, you can show that this can only happen if $a=b^2c$ for some $b\in\mathbb{Q}(i)$ and some $c\in\mathbb{Q}$ (this is the hard step; at the moment I don't see a way to prove it without using unique factorization in $\mathbb{Z}[i]$). But then $K=\mathbb{Q}(i,\sqrt{c})$ can be shown to have Galois group $\mathbb{Z}/2\times\mathbb{Z}/2$, not $\mathbb{Z}/4$.