The proof uses the non-vanishing of $L(1, \chi)$ for $\chi \neq \chi_0$.

Step 1: Character decomposition. By orthogonality: $\sum_{\substack{p \leq x \\ p \equiv a \bmod q}} \frac{1}{p} = \frac{1}{\phi(q)} \sum_{\chi \bmod q} \overline{\chi(a)} \sum_{p \leq x} \frac{\chi(p)}{p}$.

Step 2: Connection to $\log L(s, \chi)$. Taking logarithms of the Euler product: $\log L(s, \chi) = \sum_p \frac{\chi(p)}{p^s} + O(1)$ for $s > 1$, where the $O(1)$ accounts for prime powers.

Step 3: Principal character contribution. For $\chi = \chi_0$: $L(s, \chi_0) = \zeta(s)\prod_{p|q}(1 - p^{-s})$, so $\log L(s, \chi_0) \to \infty$ as $s \to 1^+$. This gives $\sum_p \chi_0(p)/p^s \to \infty$.

Step 4: Non-principal characters are bounded. For $\chi \neq \chi_0$, if $L(1, \chi) \neq 0$, then $\log L(s, \chi)$ remains bounded as $s \to 1^+$, so $\sum_p \chi(p)/p^s = O(1)$.

Step 5: Non-vanishing of $L(1, \chi)$. This is the crux. For complex $\chi$ (i.e., $\chi \neq \overline{\chi}$): consider $F(s) = \prod_\chi L(s, \chi) = \sum_{n} a_n/n^s$ where $a_n \geq 0$ (since $F(s) = \prod_\chi \prod_p (1 - \chi(p)p^{-s})^{-1} = \prod_p \prod_\chi (1 - \chi(p)p^{-s})^{-1}$, and $\prod_\chi (1 - \chi(p)x) = (1 - x^{f_p})^{\phi(q)/f_p}$ where $f_p$ is the order of $p$ in $(\mathbb{Z}/q\mathbb{Z})^\times$). Since the Dirichlet series has non-negative coefficients and a pole at $s = 1$ (from $\chi_0$), it cannot have a zero at $s = 1$ from another factor, as this would make $F(s)$ analytic at $s = 1$ with non-negative coefficients and a singularity at its abscissa of convergence -- a contradiction by Landau's theorem.

For real $\chi$: a more delicate argument using the class number formula $L(1, \chi) = \frac{2\pi h_K}{w\sqrt{|D|}}$ (for the associated quadratic field) or the product $\zeta(s)L(s,\chi) \geq 1$ (both approaches) shows $L(1, \chi) > 0$.

Step 6: Conclusion. Combining: $\sum_{\substack{p \leq x \\ p \equiv a \bmod q}} \frac{1}{p} = \frac{1}{\phi(q)}\log\log x + O(1)$, proving infinitely many primes in the progression. $\square$