This is a unit circle, obviously with a radius of 1 unit. The angle $\displaystyle \begin{align*} \theta \end{align*}$ is the angle swept out from the positive x axis in the anticlockwise direction, measured in radians. The green length is $\displaystyle \begin{align*} \cos{(\theta)} \end{align*}$, the red length is $\displaystyle \begin{align*} \sin{(\theta)} \end{align*}$ and the purple length is $\displaystyle \begin{align*} \tan{(\theta)} \end{align*}$. From the diagram, it's clear that the area of the segment is a little bigger than the area of the smaller triangle, and a little less than the area of the large triangle.

We'll deal with positive angles for the moment:

The area of the small triangle is $\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \end{align*}$, the area of the large triangle is $\displaystyle \begin{align*} \frac{1}{2}\tan{(\theta)} = \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \end{align*}$, and the area of the segment is $\displaystyle \begin{align*} \frac{\theta}{2\pi} \cdot \pi \cdot 1^2 = \frac{1}{2}\theta \end{align*}$, thus...

$\displaystyle \begin{align*} \frac{1}{2}\sin{(\theta)}\cos{(\theta)} \leq \frac{1}{2}\theta &\leq \frac{1}{2}\cdot \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \sin{(\theta)}\cos{(\theta)} \leq \theta &\leq \frac{\sin{(\theta)}}{\cos{(\theta)}} \\ \cos{(\theta)} \leq \frac{\theta}{\sin{(\theta)}} &\leq \frac{1}{\cos{(\theta)}} \\ \frac{1}{\cos{(\theta )}} \geq \frac{\sin{(\theta)}}{\theta} &\geq \cos{(\theta ) } \\ \cos{(\theta )} \leq \frac{\sin{(\theta)}}{\theta} &\leq \frac{1}{\cos{(\theta)} } \end{align*}$

And now as $\displaystyle \begin{align*} \theta \to 0, \cos{(\theta)} \to 1 \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} \to 1 \end{align*}$, thus $\displaystyle \begin{align*} \frac{\sin{(\theta)}}{\theta} \to 1 \end{align*}$ as $\displaystyle \begin{align*} \theta \to 0 \end{align*}$.

Of course this only proves the right hand limit, where you are approaching 0 from positive values of $\displaystyle \begin{align*} \theta \end{align*}$, you need to prove the left hand limit as well, where you make $\displaystyle \begin{align*} \theta \to 0 \end{align*}$ from negative values as well. But the proof of that is almost identical