Why $x<\tan{x}$ while $0<x<\frac{\pi}{2}$?

$\ \ \ \bullet$ Using similar triangles: $$\color{darkgreen}{\tan t}={\color{maroon}{\sin t}\over\color{darkblue}{\cos t}} ={ {\text{length}( \color{darkgreen} {\overline{{IZ}})} }\over 1 }\quad \Longrightarrow \quad\color{darkgreen}{\tan t}=\text{length}(\color{darkgreen}{\overline{IZ}})$$

$\ \ \ \bullet$ $t$ is the length of the arc $\color{orange}{IQ}$.

$\ \ \ \bullet$ Area of the circular sector $O\color{orange}{IQ}={t\over 2\pi}\cdot \pi\cdot 1^2={t\over2}$.

$\ \ \ \bullet$ Area of $\triangle OIZ={1\over2}\cdot1\cdot\color{darkgreen}{\tan t}$.

So $$\color{maroon}{\sin t}\lt t\lt\color{darkgreen}{\tan t}$$ for $0< t<\pi/2$.

It's fine if the comparisons of those lengths is intuitively clear to you, but if you want to be rigorous, it's easier to compare nested areas than it is to compare curve lengths.

Working off of David Mitra's picture (see his answer), the area of the triangle spanned by the lines $\cos(t)$ and $\sin(t)$ is $\frac{1}{2} \cos(t)\sin(t)$. That area rests inside the sector of angle $t$, which has area $\frac{t}{2}$ (the proportion $\frac{t}{2\pi}$ of the entire circle's area $\pi$). And in turn, the area of sector is inside the triangle spanned by the horizontal radius of the unit circle and $\tan(t)$, which has area $\frac{1}{2} \tan(t)$.

Thus we have the inequality $$\frac{1}{2} \cos(t)\sin(t) \le \frac{t}{2} \le \frac{1}{2} \tan(t)$$ Or $$\cos(t)\sin(t) \le t \le \tan(t)$$ canceling the 2's. You should be able to adjust your proof of $\displaystyle\lim_{t \to 0} \frac{\sin(t)}{t} = 1$ to utilize this slightly weaker inequality.