Note that, since $\pi/3\gt1\gt\pi/4$, we have $\sin1\gt\cos1\gt\cos(\pi/3)=1/2$. It follows that
$$\sin^2(n-1)+\sin^2(n+1)=(\sin n\cos1-\cos n\sin 1)^2+(\sin n\cos1+\cos n\sin1)^2\\
=2(\sin^2n\cos^21+\cos^2n\sin^21)\\
\gt2(\sin^2n+\cos^2n)/4=1/2$$
Therefore, taking three consecutive terms of the sum, we see that
$${|\sin(3k-1)|\over3k-1}+{|\sin3k|\over3k}+{|\sin(3k+1)|\over3k+1}\gt{\sin^2(3k-1)+\sin^2(3k+1)\over3k+1}\gt{1\over2(3k+1)}$$
(by ignoring the middle term altogether, using the general inequality $|\sin x|\ge\sin^2x$, and taking the larger of the two remaining denominators). Thus
$$\sum_{n=1}^\infty{|\sin n|\over n}\gt|\sin1|+\sum_{k=1}^\infty{1\over6k+2}=\infty$$
(Note, there was no real need to include the first term, $|\sin1|$ on the right hand side in the final inequality; I kept it just only to point out that it did not participate in any of the triples of consecutive terms, which were all centered on multiples of $3$.)