Proving the length of a circle's arc is proportional to the size of the angle

I believe this is a matter of definition. (One) Radian measure is defined as the angle subtended at the centre by an arc of length equal to the radius of a circle. So for any arc of length $l$, the angle (in radians) is $\theta = \dfrac l r$. So $ l = r\cdot\theta $

Can I prove the above? Or is it something that you should just accept?

Of course you should not "just accept" it. A full formal proof would be a bit more than I'm prepared to give, but a few rather convincing propositions give a good argument.

1. The arc length is (strictly) monotonic.
2. The arc length is invariant under rotations.

From the second point, we get

$$l(k\cdot\alpha) = k\cdot l(\alpha)$$

for $k \in \mathbb{N}$ directly, and subdividing, we get

$$l\left(\frac{k}{m}\alpha\right) = \frac{k}{m}l(\alpha)$$

for $k \in \mathbb{N},\; m \in \mathbb{Z}^+$, so the proportionality for rational multiples of the angle $\alpha$

Then, for irrational $t > 0$, we get

$$\sup \{q\cdot l(\alpha)\colon q \in \mathbb{Q}, 0 < q < t\} \leqslant l(t\cdot \alpha) \leqslant \inf\{q\cdot l(\alpha)\colon q \in \mathbb{Q}, q > t\}$$

from the monotonicity.

I am not sure if this is completely true, but I am sure others will correct me if I am mistaken.

Consider a circle of radius R. In polar form, this equation is $r(\theta)=R$. The equation for arc length is as follows:

$L = \int_0^{\theta} \sqrt {r^2+ (\frac {dr} {d\theta})^2 } d\theta$

We know that r=R and dr/d$\theta$ is 0, so the integral becomes:

$L = \int_0^{\theta} R d\theta$

L=$\theta$R