There are proofs that treats the cases of real and non-real $\chi$ on an equal footing. One proof is in Serre's Course in Arithmetic, which the answers by Pete and David are basically about. That method is using the (hidden) fact that the zeta-function of the $m$-th cyclotomic field has a simple pole at $s = 1$, just like the Riemann zeta-function.
Here is another proof which focuses only on the $L$-function of the character $\chi$ under discussion, the $L$-function of the conjugate character, and the Riemann zeta-function.

Consider the product
$$
H(s) = \zeta(s)^2L(s,\chi)L(s,\overline{\chi}).
$$
This function is analytic for $\sigma > 0$, with the
possible exception of a pole at $s = 1$. (As usual I write $s = \sigma + it$.)

Assume $L(1,\chi) = 0$. Then also $L(1,\overline{\chi}) = 0$.
So in the product defining $H(s)$, the
double pole of $\zeta(s)^2$ at $s = 1$ is
cancelled and $H(s)$ is therefore analytic throughout the half-plane
$\sigma > 0$.

For $\sigma > 1$, we have the exponential representation
$$
H(s) = \exp\left(\sum_{p, k} \frac{2 + \chi(p^k) + \overline{\chi}(p^k)}
{kp^{ks}}\right),
$$
where the sum is over $k \geq 1$ and primes $p$. If $p$ does not divide
$m$, then we write $\chi(p) = e^{i\theta_p}$ and find

$$
\frac{2 + \chi(p^k) + \overline{\chi}(p^k)}{k} =
\frac{2(1 + \cos(k\theta_p))}{k} \geq 0.
$$

If $p$ divides $m$ then this sum is $2/k > 0$.

Either way, inside that exponential is a Dirichlet series with nonnegative coefficients, so when we exponentiate and rearrange terms (on the half-plane of abs. convergence, namely where $\sigma > 1$), we see that $H(s)$ is a Dirichlet series with nonnegative coefficients. A lemma of Landau on Dirichlet series with nonnegative coefficients then assures us that *the Dirichlet series representation of $H(s)$
is valid on any half-plane where $H(s)$ can be analytically continued.*

To get a contradiction at this point, here are
several methods.

[Edit: In the answer by J.H.S., and due to Bateman, is the slickest argument I have seen, so let me put it here. The idea is to look at the coefficient of $1/p^{2s}$ in the Dirichlet series for $H(s)$.
By multiplying out the $p$-part of the Euler product, the coefficient of $1/p^s$ is $2 + \chi(p) + \overline{\chi}(p)$, which is nonnegative, but the coefficient of $1/p^{2s}$ is $(\chi(p) + \overline{\chi}(p) + 1)^2 + 1$, which is not only nonnegative but in fact is greater than or equal to 1. Therefore if $H(s)$ has an analytic continuation along the real line out to the number $\sigma$, then
for real $s \geq \sigma$ we have $H(s) \geq \sum_{p} 1/p^{2s}$.
The hypothesis that $L(1,\chi) = 0$ makes $H(s)$ analytic for all complex numbers with positive real part, so we can take $s = 1/2$ and get $H(1/2) \geq \sum_{p} 1/p$, which is absurd since that series over the primes diverges. QED!]

If you are willing to accept that
$L(s,\chi)$ (and therefore $L(s,\overline{\chi})$) has
an analytic continuation to the whole plane, or
at least out to the point $s = -2$,
then $H(s)$ extends to $s = -2$. The Dirichlet
series representation of $H(s)$ is convergent at $s = -2$ by our analytic continuation hypothesis and it shows $H(-2) > 1$, or the exponential representation implies that at least $H(-2) \not= 0$.
But $\zeta(-2) = 0$, so $H(-2) = 0$. Either way, we have a contradiction.

There is a similar argument, pointed out to me
by Adrian Barbu, that does not
require analytic continuation of $L(s,\chi)$
beyond the half-plane $\sigma > 0$. If you are willing to accept
that $\zeta(s)$ has
zeros in the critical strip $0 < \sigma < 1$ (which is a region that the Dirichlet series and exponential representations of $H(s)$ are both valid since $H(s)$ is analytic on $\sigma > 0$), we can evaluate the exponential representation of $H(s)$ at such a zero to
get a contradiction. Of course the amount of
analysis that lies behind this
is more substantial than what is used to
continue $L(s,\chi)$ out to $s = -2$.

We consider $H(s)$ as $s \rightarrow 0^{+}$. We need to accept that $H$ is bounded as $s \rightarrow 0^{+}$. (It's even
holomorphic there, but we don't quite need that.)
For real $s > 0$ and a fixed prime $p_0$ (not dividing $m$, say), we
can bound $H(s)$
from below by the sum of the $p_0$-power terms in its Dirichlet series.
The sum of these terms is exactly the $p_0$-Euler factor of $H(s)$, so we
have the lower bound
$$
H(s) >
\frac{1}{(1 - p_0^{-s})^2(1 - \chi(p_0)p_0^{-s})(1 -
\overline{\chi}(p_0)p_0^{-s})} = \frac{1}{(1 - p_0^{-s})^2(1 - (\chi(p_0)+ \overline{\chi}(p_0))p_{0}^{-s} + p_0^{-2s})}
$$
for real $s > 0$. The right side tends to $\infty$ as $s \rightarrow 0^{+}$.
We have a contradiction. QED

These three arguments at some point use knowledge beyond the half-plane $\sigma > 0$ or a nontrivial zero of the zeta-function. Granting any of those lets you see easily that $H(s)$ can't vanish at $s = 1$, but that "granting" may seem overly technical. If you want a proof for the real and complex cases uniformly which does not go outside the region $\sigma > 0$, use the method in the answer by Pete or David [edit: or use the method I edited in as the first one in this answer].

isa fundamental distinction between the two cases. The idea is, that if $L(1,\chi)=0$ for $\chi$ complex, then so does its conjugate, which gives a double zero in the product that David Speyer gives below. Enlarging this idea, we get a fairly decent zero-free region for $L(1,\chi)=0$ for $\chi$ complex (I don't recall, but $1/\log D$ maybe). But for real characters this is not true, and we have to worry about the so-called Siegel zeros, and the zero-free region is much worse (as $1/\sqrt D$). The difference is the single vs double effect. – Junkie May 25 '10 at 5:03