Corollary 4 (p. 123) in Hlawka, Taschner, and Schoissengeier, Geometric and Analytic Number Theory.
Theorem 7.10 (p. 111) in Ayoub, An Introduction to the Analytic Theory of Numbers.
Theorem 3.4.5 (p. 134) in Jameson, The Prime Number Theorem.
Theorem 5.20 (p. 239) in Narkiewicz, The Development of Prime Number Theorey from Euclid to Hardy and Littlewood.
Theorem 5.9 (p. 100) in Bateman and Diamond, Analytic Number Theory: An Introductory Course.
In the first and third references, the proof is essentially based on the following theorem. If the Dirichlet series $f(s) = \sum_{n \geq 1} b_n/n^s$ has bounded coefficients, so it converges for ${\rm Re}(s) > 1$, and $f$ has an analytic continuation to the line ${\rm Re}(s) = 1$ except for at worst a simple pole at $1$, with $$ f(s) = \frac{\rho}{s-1} + c + O(s-1) $$ near $s = 1$, then $(1/x)\sum_{n \leq x} b_n \to \rho$ and $\sum_{n \leq x} b_n/n = \rho\log x + c + o(1)$ as $x \to \infty$. In particular, if $\rho = 0$, meaning $f$ is analytic at $s = 1$ and $c = f(1)$, then $(1/x)\sum_{n \leq x} b_n \to 0$ as $x \to \infty$ and $f(1) = \sum_{n \geq 1} b_n/n$. Taking $f(s) = \sum \mu(n)/n^s = 1/\zeta(s)$, we get $(1/x)\sum_{n \leq x} \mu(n) \to 0$ and $\sum_{n \geq 1} \mu(n)/n = 0$ since $1/\zeta(s)$ is holomorphic on ${\rm Re}(s) \geq 1$ with a zero at $s = 1$.