Length of a Chord of a circle

This comes down to the intermediate value theorem.

On the circle $x^2+y^2=1$, the chord from $(1,0)$ to $(\cos\theta,\sin\theta)$ has length $2\sin\dfrac\theta2$. You can see that by means of the usual "distance formula".

As $\theta$ goes from $0$ to $\pi$ (or from $0^\circ$ to $180^\circ$ if you like), the chord goes from $0$ to $2$ and the chord is a continuous function of $\theta$. The fact that it's continuous means you can apply the intermediate value theorem and see that it assumes all intermediate values.

If you don't like transcendental functions (perhaps because proving continuity of those takes a lot of work), you can also do it like this: the point $$ \left( \frac{1-t^2}{1+t^2}, \frac{2t}{1-t^2} \right) \tag 1 $$ goes around the circle from $(-1,0)$ back to $(-1,0)$ as $t$ goes from $-\infty$ to $+\infty$. The length of the chord from $(1,0)$ to the point in $(1)$ can also be found via the distance formula and the same kind of argument can be used.

Tangentially (no pun intended) related is this: Ptolemy's table of chords

Open a compass with the desired length $x$, place the needle on the circumference and draw an arc that intersects the circle. Can you imagine that you'll never meet it ?

Hint:$$\text{ chord length } = 2r\sin\bigg({\frac{\theta}{2}}\bigg)$$ where $r$ is the radius and $\theta$ is the angle subtended at the center by the chord. Note the continuity of the RHS, now use the Intermediate Value Theorem.