I'll borrow heavily from @robjohn's answer to the question for $\lim\limits_{x\to 0}\frac{\sin{x}}{x}$, since that seems to be the natural jumping off point for this question. But let's add some new points to his diagram: we have $\angle BCE$ as a right angle by Thales's theorem, and $\angle BEC = \frac{1}{2} \angle BAC = \frac{1}{2}x$ since the angle subtended by $BE$ at the circumference is half the angle it subtends at the centre. This also means $\angle CBE = \frac{\pi}{2} - \frac{1}{2}x$ and so $\angle BCF = \frac{1}{2}x$.
Let's follow @robjohn for a while since we'll use several of his results. The area of $\triangle ABC$ is $\frac{1}{2}\sin(x)$, the area of the larger $\triangle ABD$ is $\frac{1}{2}\tan(x)$, and the area of the blue sector is $\frac{1}{2}x$. Since the latter falls between the areas of those triangles,
$$\sin(x) \le x \le \tan(x) \tag{1}$$
Dividing by $\sin(x)$ and taking reciprocals, we deduce
$$\cos(x) \le \frac{\sin(x)}{x} \le 1 \tag{2}$$
Since $\cos(x) \to 1$ as $x \to 0$, we obtain the desired "squeeze":
$$\lim_{x\to0} \frac{\sin(x)}{x} = 1 \tag{3}$$
Dividing $(2)$ by $\cos(x)$, we get that
$$1 \le \frac{\tan(x)}{x} \le \sec(x) \tag{4}$$
Since $\sec(x) \to 1$ as $x \to 0$, we get another squeeze:
$$\lim_{x\to0}\frac{\tan(x)}{x}=1\tag{5}$$
Now I seek to extend this to find a "squeezeable" bound for $(1 - \cos(x))/x$. We know $\cos(x) \le 1$ and for acute angles $0 \le x \le \frac{\pi}{2}$ we
have $\sin(x) \le x$ from $(1)$, so in this range we can bound our desired expression by
$$0 \le \frac{1 - \cos(x)}{x} \le \frac{1 - \cos(x)}{\sin(x)} = \tan \left( \frac{x}{2} \right) \tag{6}$$
Here we used the fact that this ratio appears in $\triangle BCF$. Intuitively, this triangle gets more "pointy" as the angle $x$ narrows, and so the $1 - \cos(x)$ leg goes to zero faster than the $\sin(x)$ leg does. Now since $\tan(\frac{1}{2}x) \to 0$ as $x \to 0^+$, we can squeeze:
$$\lim_{x \to 0^+} \frac{1 - \cos(x)}{x} = 0$$
Since $\tan(\frac{1}{2}x)$ and $(1 - \cos(x))/x$ are odd functions, they will also clearly approach zero as $x \to 0^-$ and so we get the desired result
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0 \tag{7}$$
A much more useful result is that
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} $$
This extends our linear approximation from $(7)$ for small angles, that $\cos(x) = 1 + O(x^2)$, to the quadratic approximation $\cos(x) = 1 - x^2/2 + O(x^3)$. By a similar argument to $(6)$,
$$\frac{1 - \cos(x)}{x^2/2} \le \frac{1 - \cos(x)}{\sin(x) \cdot x/2} = \frac{\tan(x/2)}{x/2} \tag{8}$$
From $(5)$, the limit of the right-hand side as $x \to 0$ is $1$. This is only half the squeeze; we also need a suitable lower bound. We can use the bound $x \le \tan(x)$ from $(1)$ to give
$$\frac{1 - \cos(x)}{x^2/2} \ge \frac{1 - \cos(x)}{\tan(x) \cdot x/2} = \cos(x) \cdot \frac{1 - \cos(x)}{\sin(x) \cdot x/2} = \cos(x) \cdot \frac{\tan(x/2)}{x/2} \tag{9}$$
Using $(5)$ and the fact $\cos(x) \to 1$ as $x \to 0$, this right-hand side also goes to $1$ as $x \to 0$ (limit of a product is product of the limits). So by $(8)$ and $(9)$ we now have both sides of the squeeze and obtain the desired
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x^2/2} = 1 \tag{10}$$
Here's a graph showing how the squeeze looks.
However, I feel steps $(8)$ to $(10)$ have somewhat lost touch with the underlying geometry — although they recycle the earlier, more geometrically motivated results, they largely aren't using quantities we can visualise on the diagram. Perhaps there is a more "geometric" way to get this stronger result by comparing arc lengths or areas on the diagram in a more precise manner.
An alternative kind of geometric intuition might come from viewing $(1 - \cos(x))/\sin(x)$ as $\csc(x) - \cot(x)$. If we can show that as $x \to 0$ the difference between cosec and cot also goes to zero, that would suffice to squeeze $(1 - \cos(x))/x$ to zero.
Now $\cot(x)$ is the horizontal leg of a right-angled triangle whose other leg is the vertical radius of the unit circle (so fixed at length one), and $\csc(x)$ is its hypotenuse. As $x \to 0$, this right-angled triangle gets larger, and the other acute angle becomes closer to another right angle.
In general, as you increase the size of a right-angled triangle for which one leg is fixed then the two variable sides get closer in length as your triangle becomes closer to isosceles.
A short algebraic proof: let the fixed leg be of length $1$, the other leg be $a$, and the hypotenuse $\sqrt{a^2+1}$. Then
$$\lim_{a \to \infty} \left(\sqrt{a^2+1} - a \right) = \lim_{a \to \infty} \frac{(\sqrt{a^2+1} - a)(\sqrt{a^2+1} + a)}{\sqrt{a^2+1} + a} = \lim_{a \to \infty} \frac{1}{\sqrt{a^2+1} + a} = 0.$$
But it would be nice to see this argument geometrically.
