= Solution
Fix integers $q\ge1$ and $a$ with $(a,q)=1$. We prove the <Dirichlet theorem on primes in arithmetic progressions> by showing that the <prime> Dirichlet sum in this <residue class> diverges as its real exponent decreases to one.
First justify the required <Dirichlet L-function> facts. If $\chi$ is a <nonprincipal Dirichlet character> modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi(b)\ne1$, and multiplication by $b$ permutes the <residue classes>, forcing the sum to equal $\chi(b)$ times itself. Therefore the partial sums $C_\chi(u)=\sum_{n\le u}\chi(n)$ are bounded by $q$. <Partial summation> gives
$$
L(s,\chi)=s\int_1^\infty C_\chi(u)u^{-s-1}\,du\qquad(\operatorname{Re}s>0),
$$
which supplies holomorphic continuation near one, with locally <uniform convergence>. For the <principal Dirichlet character> $\chi_0$,
$$
L(s,\chi_0)=\zeta(s)\prod_{p\mid q}(1-p^{-s})
$$
has a simple <pole> at one, with <residue> $\varphi(q)/q$.
The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square $\chi^2$ is nonprincipal. For real $\sigma>1$, logarithms of the <Euler products> give
$$
\zeta(\sigma)^3|L(\sigma,\chi)|^4|L(\sigma,\chi^2)|\ge1.
$$
For a <prime> not dividing $q$, its logarithmic coefficient is $3+4\operatorname{Re}\chi(p)^k+\operatorname{Re}\chi(p)^{2k}=2(1+\cos\theta)^2\ge0$, where $\chi(p)^k=e^{i\theta}$. <Primes> dividing $q$ contribute only the positive zeta term. If $L(1,\chi)$ vanished to order $m\ge1$, the product would be $O((\sigma-1)^{4m-3})$ and tend to zero: the squared-character factor is bounded because $\chi^2$ is nonprincipal. This is a contradiction. Hence
$$
L(1,\chi)\ne0\qquad\text{for every nonprincipal }\chi.
$$
This step uses the given real-character hypothesis and the <Euler product positivity for L-function nonvanishing> for all remaining characters.
In $\sigma>1$ choose the logarithm furnished by the absolutely convergent <Euler product>. Its prime-power expansion gives
$$
\log L(\sigma,\chi)=\sum_p\frac{\chi(p)}{p^\sigma}+O(1).
$$
The error from powers $k\ge2$ is uniformly bounded as $\sigma\downarrow1$, by $\sum_p\sum_{k\ge2}1/(kp^{k\sigma})<\infty$. For nonprincipal $\chi$, nonvanishing and holomorphic continuation at one provide a bounded logarithm in a neighborhood of one. The Euler-product branch differs from that branch by a constant integral multiple of $2\pi i$ on the final real interval, so it too is bounded. Thus
$$
\sum_p\frac{\chi(p)}{p^\sigma}=O_q(1)\qquad(\chi\ne\chi_0).
$$
For the principal character, its simple <pole> instead gives
$$
\sum_{p\nmid q}\frac1{p^\sigma}=\log\frac1{\sigma-1}+O_q(1).
$$
The <Orthogonality of Dirichlet characters> gives, for <primes> not dividing $q$,
$$
1_{p\equiv a\pmod q}=\frac1{\varphi(q)}\sum_{\chi\bmod q}\overline{\chi(a)}\chi(p).
$$
Multiply by $p^{-\sigma}$ and sum. <Absolute convergence> allows the interchange for $\sigma>1$, so
$$
\boxed{\sum_{p\equiv a\pmod q}\frac1{p^\sigma}=\frac1{\varphi(q)}\log\frac1{\sigma-1}+O_q(1).}
$$
The right side tends to infinity. If there were only finitely many <primes> in the progression, the left side would have a finite limit at one. This contradiction proves \b[infinitely many <primes> occur in every reduced <residue class>]. For $q=1$ or $2$ the character argument simply has no nonprincipal terms and gives the same conclusion.
For the final deduction the modulus $q$ is fixed. Let $A(t)=\pi(t;q,a)$. The given consequence of the <Siegel–Walfisz theorem> is $A(t)\sim t/(\varphi(q)\log t)$. <Partial summation>, including the <prime> $2$ through the lower endpoint $2^-$ when appropriate, gives
$$
\sum_{\substack{p\le x\\p\equiv a\pmod q}}\frac1p=\frac{A(x)}x+\int_2^x\frac{A(t)}{t^2}\,dt.
$$
The boundary is $O_q(1/\log x)$. Write $A(t)=t(1+\varepsilon(t))/(\varphi(q)\log t)$ with $\varepsilon(t)\to0$. The main integral is $(\log\log x-\log\log2)/\varphi(q)$. The error is $o(\log\log x)$: given $\eta>0$, choose $T$ such that $|\varepsilon(t)|\le\eta$ for $t\ge T$. Its integral from $2$ to $T$ is a fixed constant, and its absolute value from $T$ to $x$ is at most $\eta\log\log x/\varphi(q)$ plus a fixed constant. Divide by $\log\log x$ and then let $\eta\downarrow0$. We obtain the <reciprocal primes in a fixed arithmetic progression> asymptotic
$$
\boxed{\sum_{\substack{p\le x\\p\equiv a\pmod q}}\frac1p\sim\frac1{\varphi(q)}\log\log x.}
$$
The argument uses the supplied prime-counting asymptotic in this last part; the preceding infinitude proof did not assume the <Prime number theorem> in <arithmetic progressions>.
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