Shortest way of proving that the Galois conjugate of a character is still a character
OK, I've thought about it a bit more and I'll give an alternate proof below the first horizontal line. However, I highly deny that this proof is better, and I am not sure it is actually shorter. To my mind, the morally correct proof is to show that, if $K \subseteq L$ with $L$ algebraically closed, and a system of polynomial equations has a root in $L$, then it has a root in a finite extension of $K$.
The point, which I am sure Qiaochu understands, is that he only knows a priori that the representation is defined over $\mathbb{C}$. Once he knows that the representation is definable over an algebraic extension $K'$ of $K$, he can replace $K'$ by its normal closure, note that $\mathrm{Gal}(K', \mathbb{Q}) \to \mathrm{Gal}(K, \mathbb{Q})$ is surjective, lift any element $\sigma$ of $\mathrm{Gal}(K, \mathbb{Q})$ to some $\tilde{\sigma}$ in $\mathrm{Gal}(K', \mathbb{Q})$, and apply $\tilde{\sigma}$ to the entries of his matrices.
Part of the problem is that the representation may honestly not be defined over $K$. For example, the two dimensional representation of the quaternion eight group has character with values in $\mathbb{Q}$, but can't be defined over $\mathbb{Q}$.
Fix $G$ and a representation $V$ of $G$. For $g \in G$, let $\lambda_1(g)$, $\lambda_2(g)$, ..., $\lambda_n(g)$ be the multiset of eigenvalues of $g$ acting on $V$. These are necessarily roots of unity, since $g^N=1$ for some $N$. For any symmetric polynomial $f$, with integer coefficients, define $\chi(f,g) = f(\lambda_1(g), \ldots, \lambda_n(g))$.
Lemma: With notation as above, $g \mapsto \chi(f,g)$ is a virtual character.
Proof: If $f$ is the elementary symmetric function $e_k$, then this is the character of $\bigwedge^k V$. Any symmetric function is a polynomial (with integer coefficients) in the $e_k$'s; take the corresponding tensor product and formal difference of virtual characters.
Any Galois symmetry $\sigma$ of $\mathbb{Q}(\zeta_N)$ is of the form $\zeta_N \mapsto \zeta_N^s$, for $s$ relatively prime to $N$. Consider the power sum symmetric function $p_s := \sum x_i^s$. So $\chi(p_s, \ )$ is the Galois conjugate $\chi^{\sigma}$, and we now know that it is a virtual character.
But $\langle \chi^{\sigma}, \chi^{\sigma} \rangle = \langle \chi, \chi \rangle =1$, because the inner product is built out of polynomial operations and complex conjugation, and complex conjugation is central in the Galois group. So this virtual character must correspond to $\pm W$, for some representation $W$. Since $\chi^{\sigma}(e) = \chi(e) = \dim V$, we conclude that the positive sign is correct.
It just occurred to me that actually writing this out for some specific small values of $s$ makes some nonobvious statements about representation theory. For example, if $G$ has odd order and $V$ is a $G$-irrep, then $\bigwedge^2 V$ has a $G$-equivariant injection into $\mathrm{Sym}^2 V$. Proof: The difference of their characters is the character of $V^{\sigma}$, where $\sigma: \zeta \mapsto \zeta^2$.
I think the natural proof (although I think that this is the one you want to avoid) is that $\mathbb Q[G]$ is a semi-simple $\mathbb Q$-algebra, hence when extended to $\overline{\mathbb Q}$ it splits as a product of matrix rings, hence it already so splits when extended over a finite extension $K$ of $\mathbb Q$, hence all representations of $G$ are defined over this $K$. (This $K$ is not the $K$ in the statement of your question, but rather the finite extension of it that you allude to.)
The reason I am writing this out despite your request not to is just to point out that it is not that annoying to prove; in fact it is quite natural. And all the arguments that I know of the type that certains systems of eigenvalues are closed under Galois conjugation (e.g. that if $f$ is a modular form which is a Hecke eigenform with system of eigenvalues $(a_p)$, then for any Galois element $\sigma$, the system of eigenvalues $(\sigma(a_p))$ is also attached to a Hecke eigenform) are proved in the same manner. (Namely, by showing that the natural $\mathbb C$-algebra that governs the situation, whether it be the group algebra or the Hecke algebra, actually has a model over $\mathbb Q$, with the same set of generators (group elements or Hecke operators, as the case may be).)
Well, wait a minute. I guess we can agree that every complex representation $\rho$ of $G$ is defined over $\mathbb{C}$! Let $\sigma$ be an automorphism of the character field $K$. Then $\sigma$ extends to an automorphism of $\mathbb{C}$: for a proof, see for instance $\S 10$ of my field theory notes (the numbering of the results and the sections is not stable as yet, so search for automorphism extension theorem to find it quickly.) The map $g \mapsto \sigma( \rho(g))$ is clearly a representation of $G$ with character $\sigma(\chi)$, where $\chi$ is the character of $\rho$. Aren't we done?