We're on the plane $\mathbb C$ or $\mathbb R^2$.
Definition 1: A movement is a colection of isometries $H_t$ in the plane, with $t \in [0,1]$ and $H_0$ being the identity isometry and such that $H_t(X)$ is a continuous funtion on $t$.
Definition 2: A rigid body is any subset $F_t \subset \mathbb C$ which will be subjected to $H_t$.
I can prove that the most generic proper isometry that can happen in the plane is a rotation around some point (or a pure translation but that's easier I guess). Meaning that for all $H_t$, there exists a point $C_t$ such that $H_t$ is a rotation around $C_t$ of the original $H_0$.
Because $H_t$ is continuous in $t$, there exists the point $I = \lim _{t \rightarrow 0^+} C_t$ which we will call the instantaneous center of rotation at the moment $0$.
From this definition is there a good way to prove that the velocity of point $I$ must be zero at the moment $t =0$?
EDIT: position of a point $X$ is just $\vec H_t(X)$ (measured with the point $(0,0)$ which is fixed in time, because we measure speed with respect to an stationary earth and not with respect to moving points of the body, right?), velocity of a point $X$ is $\frac{d \vec H_t(X)}{dt}$.
EDIT: We should have $H_t (z)= \alpha _ t z + \beta _ t$, $|\alpha_t| = 1$, $\alpha_0=1$ and $\beta _ 0 =0$ but this seems to imply $I = (0,0)$ always.
In complex notation, as you noticed, $H_t(z)=\alpha_t z+\beta_t$. By definition, the center of rotation is the unique solution $c_t$ of $H_t(c_t)=c_t$. That is: $$c_t=\frac{\beta_t}{1-\alpha_t}$$ And the instantaneous center $I$, if it exists, is represented by complex number $$c=\lim_{t\rightarrow 0}\frac{\beta_t}{1-\alpha_t}\tag{1}$$ First, notice that the hypotheses of continuity on $H_t$, that is, on $\alpha_t$ and $\beta_t$, are not enough to guarantee the existence of that limit. For instance, if $\alpha_t=e^{it}$ and $\beta_t=\sqrt t$ the limit is at one complex infinity and the instantaneous center is at some infinity (in the direction $\frac \pi 4$) in the plane. That's because the shift/glide factor is much larger than the rotation factor, so you would expect the center to be sent to an infinity. In fact a more extreme example is what $\alpha_t=1$, which means your move is a pure shift/translation (there's no center of rotation).
But let's assume that the limit in $(1)$ exists. Then it's not necessarily the origin, as you had initially thought. It can literally be any point in the plane. More precisely, we can design $\alpha_t$ and $\beta_t$ such that the limit is any given point.
Now, by your definition, the velocity of any point $z$ at $t=0$ is given by $$\left.\frac{dH_t z}{dt}\right|_{t=0} = \left.\frac{d\alpha_t}{dt}\right|_{t=0}\cdot z + \left.\frac{d\beta_t}{dt}\right|_{t=0}\tag{2}$$
Now as you know, $\alpha_0=1$ and $\beta_0=0$, thus $$\left\{\begin{split} \left.\frac{d\alpha_t}{dt}\right|_{t=0} &= \lim_{t\rightarrow 0}\frac{\alpha_t-1}{t}\\ \left.\frac{d\beta_t}{dt}\right|_{t=0} &= \lim_{t\rightarrow 0}\frac{\beta_t}{t} \end{split}\right.\tag{3}$$ Now combining $(1)$, $(2)$ and $(3)$ gives the velocity of the instantaneous center at $t=0$ as $$\begin{split} \left.\frac{dH_t c}{dt}\right|_{t=0} &= \left.\frac{d\alpha_t}{dt}\right|_{t=0}\cdot \lim_{t\rightarrow 0}\frac{\beta_t}{1-\alpha_t} + \left.\frac{d\beta_t}{dt}\right|_{t=0}\\ &= \lim_{t\rightarrow 0}\frac{\alpha_t-1}{t} \cdot \lim_{t\rightarrow 0}\frac{\beta_t}{1-\alpha_t} + \lim_{t\rightarrow 0}\frac{\beta_t}{t}\\ &= \lim_{t\rightarrow 0}\left( \frac{\alpha_t-1}{t} \cdot \frac{\beta_t}{1-\alpha_t}+\frac{\beta_t}{t}\right)\\ &=0 \end{split}$$ which is the desired result.