Prove that $\sin nx\le n\sin x$, by Induction
Solution 1:
You also have that $\sin x\ge0$ and from $\cos kx\le1$ you conclude that $$ \sin x\cos kx\le\sin x $$ So this is the easy step.
The issue is to prove $\sin kx\cos x\le k\sin x$, where your argument is faulty. But this is not a real problem: if $\sin kx\le0$, then $\sin kx\cos x\le0\le k\sin x$. If $\sin kx>0$, then from $\cos x\le 1$ you obtain $\sin kx\cos x\le \sin kx$ and you can apply the induction hypothesis.