= Solution
For a positive integer $N$, a <Dirichlet character> $\psi$ modulo $N$ is a <group homomorphism> $(\mathbb Z/N\mathbb Z)^\times\to\mathbb C^\times$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are <roots of unity>. For $\operatorname{Re}s>1$, its <Dirichlet L-function> is the absolutely convergent <Dirichlet series>
$$
\boxed{L_N(\psi,s)=\sum_{n=1}^\infty\frac{\psi(n)}{n^s}=\prod_{p\nmid N}(1-\psi(p)p^{-s})^{-1}.}
$$
Complete multiplicativity of $\psi$ proves the <Euler product> by expanding each factor as a <geometric series> and applying unique <integer factorization>; absolute convergence justifies these rearrangements.
Suppose $\psi$ is a <nonprincipal Dirichlet character>. Choose a unit $u$ with $\psi(u)\ne1$. Multiplication by $u$ permutes a complete residue system, so $\sum_{a=1}^N\psi(a)=\psi(u)\sum_{a=1}^N\psi(a)$, and the sum is zero. Consequently $A(x)=\sum_{n\le x}\psi(n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The <Abel summation formula> gives, initially for $\operatorname{Re}s>1$,
$$
L_N(\psi,s)=s\int_1^\infty A(x)x^{-s-1}\,dx.
$$
For $\operatorname{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\operatorname{Re}s\ge\delta>0$ the differentiated integrands are bounded by constants times $x^{-1-\delta}(1+\log x)^k$. Thus <holomorphy of a Dirichlet series from bounded partial sums> proves
$$
\boxed{L_N(\psi,s)\text{ is holomorphic on }\operatorname{Re}s>0.}
$$
For a nonreal $\psi$, the character $\psi^2$ is also nonprincipal. The supplied logarithmic <Euler product> yields, for real $\sigma>1$,
$$
\log\bigl(\zeta(\sigma)^3|L_N(\psi,\sigma)|^4|L_N(\psi^2,\sigma)|\bigr)=\sum_p\sum_{k\ge1}\frac{3+4\operatorname{Re}(\psi(p)^k)+\operatorname{Re}(\psi(p)^{2k})}{kp^{k\sigma}}.
$$
At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi(p)^k=e^{i\theta}$; then the numerator is $3+4\cos\theta+\cos2\theta=2(1+\cos\theta)^2\ge0$. This <Euler product positivity for L-function nonvanishing> proves that the product inside the logarithm is at least one. If $L_N(\psi,1)=0$, its holomorphy would give $L_N(\psi,\sigma)=O(\sigma-1)$. Meanwhile $L_N(\psi^2,\sigma)$ is bounded near one and the <Riemann zeta function> has a simple <pole> there. The product would therefore be $O(\sigma-1)$ and tend to zero, a contradiction.
For a real nonprincipal $\psi$, suppose instead that $L_N(\psi,1)=0$. The product $H(s)=\zeta(s)L_N(\psi,s)$ would then be holomorphic on the whole half-plane $\operatorname{Re}s>0$, because the zero cancels the only zeta pole there. In $\operatorname{Re}s>1$ it has coefficients
$$
H(s)=\sum_{n\ge1}b_n n^{-s},\qquad b_n=\sum_{d\mid n}\psi(d).
$$
These are <nonnegative zeta-times-real-L coefficients>. They are multiplicative, and their prime-power values are $k+1$ if $\psi(p)=1$, one or zero according as $k$ is even or odd if $\psi(p)=-1$, and one if $\psi(p)=0$. In particular $b_n\ge0$ and $b_{m^2}\ge1$ for every $m$.
The <Taylor series> of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent <Dirichlet series> at $2$ gives $(-1)^kH^{(k)}(2)=\sum_n b_n(\log n)^k/n^2$. Evaluate the <Taylor series> at $s=1/2$ and apply the <Tonelli theorem> to the resulting nonnegative terms:
$$
H(1/2)=\sum_{k\ge0}\frac{(3/2)^k}{k!}\sum_{n\ge1}\frac{b_n(\log n)^k}{n^2}=\sum_{n\ge1}\frac{b_n}{n^{1/2}}\ge\sum_{m\ge1}\frac1m=\infty.
$$
This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the <Landau theorem for a Dirichlet series with nonnegative coefficients>, with the needed step proved explicitly. Combining the real and nonreal cases gives the <nonvanishing of a nonprincipal Dirichlet L-function at one>:
$$
\boxed{L_N(\psi,1)\ne0\quad\text{for every nonprincipal }\psi.}
$$
Now let $a$ be coprime to $N$. The <Dirichlet theorem on primes in arithmetic progressions> asserts that infinitely many <prime numbers> satisfy $p\equiv a\pmod N$. To prove it, separate the first-power terms in the given logarithmic <Euler product>:
$$
\log L_N(\psi,\sigma)=\sum_{p\nmid N}\frac{\psi(p)}{p^\sigma}+R_\psi(\sigma),\qquad \sigma>1.
$$
For $\sigma\ge1$, the remainder is bounded uniformly by
$$
|R_\psi(\sigma)|\le\sum_p\sum_{k\ge2}p^{-k}\le\sum_{n=2}^\infty\frac1{n(n-1)}=1.
$$
For a nonprincipal character, holomorphy and nonvanishing near one give a local <holomorphic logarithm>. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the <prime-character sum near one> is bounded:
$$
\sum_{p\nmid N}\psi(p)p^{-\sigma}=O(1),\qquad \psi\ne\psi_0.
$$
For the <principal Dirichlet character> $\psi_0$,
$$
L_N(\psi_0,\sigma)=\zeta(\sigma)\prod_{p\mid N}(1-p^{-\sigma}),\qquad \sum_{p\nmid N}p^{-\sigma}=\log\frac1{\sigma-1}+O(1).
$$
The latter follows from the simple zeta pole and the bounded remainder. Finally <Orthogonality of Dirichlet characters> gives
$$
\sum_{\substack{p\text{ prime}\\p\equiv a\pmod N}}p^{-\sigma}=\frac1{\varphi(N)}\sum_{\psi\bmod N}\overline{\psi(a)}\sum_{p\nmid N}\psi(p)p^{-\sigma}=\frac1{\varphi(N)}\log\frac1{\sigma-1}+O(1).
$$
Here $\varphi$ is the <Euler totient function>. The right-hand side tends to infinity as $\sigma\downarrow1$, whereas a finite set of primes would give a bounded left-hand side. Therefore
$$
\boxed{\gcd(a,N)=1\implies\text{infinitely many primes }p\equiv a\pmod N.}
$$
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