Why is an arbitrary subfield of a CM field totally real or a CM field? Let $L$ be a CM number field: i.e. $L$ is a number field with an automorphism $\tau$ such that for any embedding $j \colon L \hookrightarrow \mathbb{C}$, $j(\tau(\ell)) = \overline{j(\ell)}$ for any $\ell \in L$, with $z \mapsto \overline{z}$ denoting complex conjugation. 
If $F \subseteq L$ is a subfield, why must it either be totally real or CM? 
If $F$ has a real embedding $j \colon F \hookrightarrow \mathbb{R}$, we may extend $j$ to a complex embedding of $L$, and we have $j(f) = \overline{j(f)} = j(\tau(f))$ for any $f \in F$, so $\tau$ must fix $F$, and therefore $F \subseteq L_+$, where $L_+$ is the fixed field of $\tau$, i.e. the maximal totally real subfield of $L$. Therefore, $F$ is totally real. 
So now, we must prove that if $F \subseteq L$ is totally imaginary, then $F$ is actually CM. If $F$ is fixed by $\tau$ (e.g. if $F/\mathbb{Q}$ is Galois), then $\tau|_F$ gives the required involution of $F$. But I don't know how to avoid this assumption (or whether it always holds). 
 A: If $M/\mathbb{Q}$ is Galois with Galois group $G$, then a choice of embedding of $v: M \rightarrow \mathbb{C}$ gives an element $c_v \in G$ given by complex conjugation on the image. The element $c_v$ is usually called complex conjugation, and it is defined up to conjugacy --- one has $c_{\sigma v} = \sigma c_v \sigma^{-1}$.


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*A Galois extension $M/\mathbb{Q}$ is CM iff (any) complex conjugation is normal. Proof: by definition, $\tau = c_{\sigma v} = \sigma c_v \sigma^{-1}$ for all $\sigma \in G$.

*The Galois closure $M$ of a CM field $L$ is CM. Proof: it suffices to show that $c_v \in \mathrm{Gal}(M/\mathbb{Q})$ does not depend on the choice of embedding $v$, or that $c_v c^{-1}_w$ is trivial for any two embeddings $v$ and $w$. By definition, the action of $c_v$ for all infinite places $v$ on $M$ restricts to $\tau$ on $L$. It follows that $\eta:=c_v c^{-1}_w$ acts trivially on $L$ for any pair of places $v$ and $w$. To show $\eta$ is trivial, it suffices to show that every conjugate of $\eta$ fixes $L$. This will mean that $\eta$ fixes every conjugate of $L$, and thus also fixes the Galois closure $M$ of $L$, which is the compositum of all conjugates of $L$. But $\sigma \eta \sigma^{-1} = c_{\sigma v} c^{-1}_{\sigma w}$ also fixes $L$, so we are done.

*Any subfield of a CM field is CM. WLOG, by 2, assume that $F$ is a subfield of a Galois CM field M. Let $H = \mathrm{Gal}(M/F)$. It suffices to show that $c$ preserves $F$, that is, $cF = F$. For then $c$ will give an automorphism of $F$ which (by comparing with $M$) must be complex conjugation for every embedding. On the other hand, because $\langle c \rangle$ is normal, it follows that the fixed field of $cF$ is $cHc^{-1} = H$. Hence, by the Galois correspondence, $cF = M^{H} = F$.
Note that $F$ need not be Galois, e.g. take $F^{+}$ to be a real quadratic field, and $F$ to be $F^{+}(\sqrt{\alpha})$ for a totally negative element $\alpha$. Generically, the Galois closure of $F$ will have Galois group $D$, the dihedral group of order $8$.
