Here's another proof:
Let $L/F$ be a cyclic extension of degree $4$, in characteristic $\neq 2$. Let $K$ be the intermediate extension of degree $2$. Then I claim there is an element $u$ of $K$ with $N_{K/F}(u)=-1$.
Proof of Claim: Let $\sigma$ be a generator for $\mathrm{Gal}(L/F)$. Let $L=K(\sqrt{b})$; choose a particular square root of $b$, in $L$, to denote by $\sqrt{b}$. Let $\sigma(\sqrt{b}) = u \sqrt{b}$. Note that $- \sqrt{b} = \sigma^2 (\sqrt{b}) = u \sigma(u) \sqrt{b}$, so $u \sigma(u) = -1$. Applying $\sigma$ to this last relation, we also have $\sigma(u) \sigma^2(u) = -1$ so $u = \sigma^2(u)$ and $u$ is in $K$. Then $u \sigma(u) = N_{K/F}(u) = -1$. QED
Why is this relevant? It shows that we can't have a real place of $F$ become complex in $K$ as, otherwise, all norms from $K$ to $F$ would be positive. In particular, let $L/\mathbb{Q}$ has Galois group the quaternion $8$ group. Let $K$ be the biquadratic subfield of $L$, with $F_1$, $F_2$ and $F_3$ the quadratic subfields of $K$. At least one of the $F_i$ must be real. If $K$ is not totally real, then the tower $L/K/F_i$ violates the claim.
PS: In fact, every cyclic degree $4$ extension $L/F$ is of the form $F(\sqrt{a}, \sqrt{ub})$ where $a$ and $b$ are in $F$ and $u \in F(\sqrt{a})$ has norm $-1$.