How to show $ \Big\vert \frac{\sin(x)}{x} \Big\vert $ is bounded by $1$? This may be a silly question, but I cannot figure it out. I want to prove that
$ \Big\vert \frac{\sin(x)}{x} \Big\vert \leq 1 $ for $x\in[-1,0)\cup(0,1]$,
but I don't even know where to start.

Simple geometric proof:
Consider the  unit circle, centred at $O$, with origin $A(1,0)$, and the point $M$ on the unit circle such that $\overset{\displaystyle\frown}{AM}=x$. Then
$$\text{triangle } OAM \subset \text{circle sector }OAM  $$ 
hence
$$\text{area tr. }OAM=\frac12\lvert\sin x\rvert\le \text{area c. sect.}OAM=\frac12\lvert x\rvert.$$
