The fundamental theorem of Galois theory says that for a finite Galois extension $L/F$ with Galois group $G=\mathrm{Gal}(L/F)$, there is an order-reversing, degree-preserving bijection
$$\{\text{intermediate fields }L\supseteq E\supseteq F\}\longleftrightarrow\{\text{subgroups }G\supseteq H\supseteq\{e\}\}$$
where the map from left to right is
$$E\longmapsto \mathrm{Fix}(E)=\{\sigma\in G:\sigma|_E=\mathrm{id}_E\}$$
If $L/\mathbb{Q}$ is a cyclic extension of degree divisible by $4$, this means $L/\mathbb{Q}$ is a Galois extension with
$$\mathrm{Gal}(L/\mathbb{Q})\cong C_n,\qquad 4\mid n$$
One basic property of $C_n$ is that for any $d\mid n$, the number of elements of order $d$ is precisely $\varphi(d)$, where $\varphi$ is Euler's totient function. In particular there is only one element of order $2$ in $\mathrm{Gal}(L/\mathbb{Q})$.
Because $a<0$ the number $\sqrt{a}$ is imaginary, and $\sqrt{a}\in L$ so that complex conjugation $\rho:L\to L$ is among the elements of $\mathrm{Gal}(L/\mathbb{Q})$. Obviously $\rho$ is an element of order $2$, so it must be the unique such element. However, the intermediate field
$$L\supseteq\mathbb{Q}(\sqrt{a})\supseteq\mathbb{Q}$$
has a corresponding subgroup $\mathrm{Gal}(L/\mathbb{Q})\supseteq H\supseteq\{0\}$ with
$$|H|=[L:\mathbb{Q}(\sqrt{a})]=\frac{n}{2}$$
Because $n$ is divisible by $4$, this is still even, and $H$ is a cyclic group (because it is a subgroup of a cyclic group) so that $H$ contains an element of order $2$. Thus it must contain $\rho$, the only element of order $2$ in $\mathrm{Gal}(L/\mathbb{Q})$. But
$$H=\{\sigma\in \mathrm{Gal}(L/\mathbb{Q}):\sigma|_{\mathbb{Q}(\sqrt{a})}=\mathrm{id}_{\mathbb{Q}(\sqrt{a})}\}$$
cannot contain $\rho$ because complex conjugation does not fix $\sqrt{a}$; contradiction.
