[proofplan]
The statement is restricted to the nonreal-valued case, which is the genuinely complex-character case. We first verify that $\chi^2$ is still nonprincipal, and then study the auxiliary Euler product $G(s)=\zeta(s)^3L(s,\chi)^4L(s,\chi^2)$ for real $s>1$. Its logarithm has a prime expansion whose first-order coefficients are nonnegative, giving a uniform lower bound for $\log |G(s)|$ as $s \to 1^+$. If $L(1,\chi)=0$, then the zero of $L(s,\chi)^4$ dominates the pole of $\zeta(s)^3$, while $L(s,\chi^2)$ stays bounded, forcing $G(s)\to 0$ and hence $\log |G(s)|\to -\infty$, a contradiction.
[/proofplan]
[step:Verify that the square of the nonreal character is nonprincipal]
Since the theorem statement assumes that $\chi$ is not real-valued, no reduction to the real-character case is needed.
Define the Dirichlet character
\begin{align*}
\psi: \mathbb{Z} &\to \mathbb{C} \\
n &\mapsto \chi(n)^2.
\end{align*}
We claim that $\psi$ is nonprincipal. If $\psi$ were principal modulo $q$, then for every integer $n$ with $\gcd(n,q)=1$ we would have $\chi(n)^2=1$, hence $\chi(n)\in\{-1,1\}$. For $\gcd(n,q)>1$ we have $\chi(n)=0$. Thus $\chi$ would be real-valued on all of $\mathbb{Z}$, contradicting the standing assumption. Hence $\chi^2=\psi$ is nonprincipal.
[/step]
[step:Record boundedness of the nonprincipal $L$-functions near $s=1$]
Let $\rho: \mathbb{Z}\to \mathbb{C}$ be any nonprincipal Dirichlet character modulo $q$. Define the partial-sum map
\begin{align*}
A_\rho: \mathbb{N} &\to \mathbb{C} \\
N &\mapsto \sum_{n=1}^{N} \rho(n).
\end{align*}
For $\operatorname{Re}(s)>1$, define the Dirichlet $L$-function associated to $\rho$ by
\begin{align*}
L_\rho: \{s\in\mathbb{C}:\operatorname{Re}(s)>1\} &\to \mathbb{C} \\
s &\mapsto \sum_{n=1}^{\infty}\frac{\rho(n)}{n^s},
\end{align*}
and write $L(s,\rho):=L_\rho(s)$. Since $\rho$ is periodic modulo $q$ and nonprincipal, the sum over one complete residue system is zero:
\begin{align*}
\sum_{r=1}^{q} \rho(r)=0.
\end{align*}
Therefore $|A_\rho(N)|\le q$ for every $N\in\mathbb{N}$.
For $N\in\mathbb{N}$ and complex $s$ with $\operatorname{Re}(s)>1$, summation by parts gives
\begin{align*}
\sum_{n=1}^{N} \frac{\rho(n)}{n^s}
=
A_\rho(N)N^{-s}
+
\sum_{n=1}^{N-1} A_\rho(n)\left(n^{-s}-(n+1)^{-s}\right).
\end{align*}
On every compact set $K\subset\{s\in\mathbb{C}:\operatorname{Re}(s)>0\}$, the terms on the right converge uniformly as $N\to\infty$, because $|A_\rho(n)|\le q$ and
\begin{align*}
\left|n^{-s}-(n+1)^{-s}\right|
\le C_K n^{-1-\sigma_K},
\qquad
\sigma_K:=\inf_{s\in K}\operatorname{Re}(s)>0,
\end{align*}
where $C_K>0$ is a constant depending only on $K$. Thus $L(s,\rho)$ extends holomorphically to $\operatorname{Re}(s)>0$, and in particular it is bounded on some closed disc centered at $1$.
Applying this to $\rho=\chi$ and $\rho=\chi^2$, both $L(s,\chi)$ and $L(s,\chi^2)$ are holomorphic and bounded in a neighbourhood of $s=1$.
[/step]
[step:Expand the logarithm of the auxiliary Euler product]
Let $\zeta: \{s\in\mathbb{C}:\operatorname{Re}(s)>1\}\to\mathbb{C}$ denote the Riemann zeta function defined by
\begin{align*}
\zeta(s):=\sum_{n=1}^{\infty}\frac{1}{n^s}.
\end{align*}
For real $s>1$, define the auxiliary function
\begin{align*}
G: (1,\infty) &\to \mathbb{C} \\
s &\mapsto \zeta(s)^3 L(s,\chi)^4 L(s,\chi^2).
\end{align*}
The absolutely convergent Euler products for $\zeta(s)$, $L(s,\chi)$, and $L(s,\chi^2)$ give
\begin{align*}
G(s)
=
\prod_{p}
(1-p^{-s})^{-3}
(1-\chi(p)p^{-s})^{-4}
(1-\chi(p)^2p^{-s})^{-1},
\end{align*}
where the product is over all primes $p$.
For real $s>1$, each Euler factor above is nonzero because $|p^{-s}|<1$, $|\chi(p)p^{-s}|<1$, and $|\chi(p)^2p^{-s}|<1$. The Euler products are absolutely convergent, so their logarithms may be expanded termwise using the logarithmic series
\begin{align*}
-\log(1-z)=\sum_{m=1}^{\infty}\frac{z^m}{m}, \qquad |z|<1.
\end{align*}
Applying this with $z=p^{-s}$, $z=\chi(p)p^{-s}$, and $z=\chi(p)^2p^{-s}$, absolute convergence permits summing over $p$ and $m$, and taking real parts yields
\begin{align*}
\log |G(s)|
=
\sum_{p}\sum_{m=1}^{\infty}
\frac{3+4\operatorname{Re}(\chi(p)^m)+\operatorname{Re}(\chi(p)^{2m})}{m p^{ms}}.
\end{align*}
Separate the $m=1$ terms from the terms with $m\ge 2$. Since $|\chi(p)|\le 1$, the latter contribution is bounded in absolute value, uniformly for $1<s\le 2$, by
\begin{align*}
8\sum_{p}\sum_{m=2}^{\infty}\frac{1}{p^m}
\le
8\sum_{n=2}^{\infty}\sum_{m=2}^{\infty}\frac{1}{n^m}
<\infty.
\end{align*}
Hence
\begin{align*}
\log |G(s)|
=
\sum_p \frac{3+4\operatorname{Re}\chi(p)+\operatorname{Re}(\chi(p)^2)}{p^s}
+
O(1)
\end{align*}
as $s\to 1^+$.
[/step]
[step:Use positivity of the first prime coefficients to bound $\log |G(s)|$ below]
For a prime $p$ with $p\nmid q$, the value $\chi(p)$ has modulus $1$, so there exists $\theta_p\in\mathbb{R}$ such that
\begin{align*}
\chi(p)=e^{i\theta_p}.
\end{align*}
Then
\begin{align*}
3+4\operatorname{Re}\chi(p)+\operatorname{Re}(\chi(p)^2)
&=
3+4\cos\theta_p+\cos(2\theta_p)\\
&=
3+4\cos\theta_p+2\cos^2\theta_p-1\\
&=
2(1+\cos\theta_p)^2
\ge 0.
\end{align*}
If $p\mid q$, then $\chi(p)=0$, so the same coefficient equals $3\ge 0$. Therefore every first-order prime coefficient is nonnegative.
The expansion from the previous step gives
\begin{align*}
\log |G(s)| \ge -C
\end{align*}
for all real $s$ sufficiently close to $1$ with $s>1$, where $C>0$ is a constant coming from the uniformly bounded $m\ge 2$ part of the logarithmic expansion.
[/step]
[step:Derive the contradiction from a hypothetical zero of $L(s,\chi)$ at $s=1$]
Assume for contradiction that
\begin{align*}
L(1,\chi)=0.
\end{align*}
Since $L(s,\chi)$ is holomorphic near $1$, there exists a function $h$ holomorphic and bounded near $1$ such that
\begin{align*}
L(s,\chi)=(s-1)h(s).
\end{align*}
Also $L(s,\chi^2)$ is bounded near $1$ because $\chi^2$ is nonprincipal.
Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. For real $1<s<2$, the elementary integral comparison gives
\begin{align*}
\zeta(s)
=
\sum_{n=1}^{\infty}\frac{1}{n^s}
\le
1+\int_1^\infty t^{-s}\,d\mathcal{L}^1(t)
=
1+\frac{1}{s-1}.
\end{align*}
Since $0<s-1<1$, this implies $\zeta(s)\le 2(s-1)^{-1}$. Hence, with $C_1:=8$,
\begin{align*}
|\zeta(s)|^3 \le C_1(s-1)^{-3}
\end{align*}
for all real $1<s<2$. Since $h$ and $L(s,\chi^2)$ are bounded near $1$, there is a constant $C_2>0$ such that
\begin{align*}
|L(s,\chi)^4 L(s,\chi^2)|
\le
C_2(s-1)^4
\end{align*}
for all real $s>1$ sufficiently close to $1$. Combining these estimates,
\begin{align*}
|G(s)|
=
|\zeta(s)|^3 |L(s,\chi)|^4 |L(s,\chi^2)|
\le
C_1C_2(s-1).
\end{align*}
Hence $G(s)\to 0$ as $s\to 1^+$, so
\begin{align*}
\log |G(s)|\to -\infty.
\end{align*}
This contradicts the lower bound $\log |G(s)|\ge -C$ obtained from the Euler-product logarithm. Therefore the assumption $L(1,\chi)=0$ is false, and
\begin{align*}
L(1,\chi)\ne 0.
\end{align*}
[/step]
