Assume, for some $\epsilon$ small enough, we have $\sin(\epsilon)\le\epsilon$. (I'll deal with this statement later.)
Theorem: Assuming the above statement, we have $\sin(a\epsilon)\le a\epsilon$ for all $a\ge1$.
We only need to prove it for $a\epsilon\le1$, because we know that $\sin 1\le1$ (because $\sin$ is always less then 1). We proceed by induction. We know (assuming the above statement) that it's true for $a=1$. Now, assume it's true for $a$; we need to prove it for $a+1$.
Note that, for $0<a\epsilon<1$, we have:
and$$0<\sin(a\epsilon)\le a\epsilon,\ 0<\cos(\epsilon)\le1$$
(The inequalities with sine follow from the hypothesis and the induction hypothesis.)
Multiplying them together, we have:
(We needed to know that they were positive, because then we know we don't have to switch around the inequality.)
Adding them together:
where I used the sum formula for sine in the last line. QED.
Now, here I'm going to have to use some sketchiness. Recall how, with radians, $\sin \epsilon\approx\epsilon$ when $\epsilon$ is small. Thus, if we let $\epsilon$ be an infinitesimal number (I told you I'm going to have to use some sketchiness), we basically have $\sin\epsilon=\epsilon$. Now, because $\epsilon$ is infinitesimal, every real number $x$ is a multiple of it. So, using the above theorem, we now have $\sin x\le x$ for all positive $x$. (A sketchy) QED.
If anything in this comment is incoherent, I apologize—I am currently extremely tired.