Does $a_n = n\sin n$ have a convergent subsequence?
Yes.
Since $\pi$ is irrational, we have infinitely many rationals $p/q$ (Dirichlet's approximation theorem) such that $$\left| \pi - \frac{p}{q} \right | < \frac{1}{q^2}$$ This implies $$|\sin p | = |\sin(q\pi - p) | < |q\pi - p| < \frac{1}{q} $$ Hence $|p\sin p| < \frac{p}{q}$, which is bounded.
Thus we have obtained a bounded subsequence of $\{n\sin n\}$, from which we can further select a convergent subsequence.