I have been trying to reach origin $(0, 0, 0)$ with the so-called Brockett integrator.
$$ \begin{bmatrix} \dot x \\ \dot y \\ \dot z \end{bmatrix} = \begin{bmatrix} u \\ v \\ yu - xv \end{bmatrix} \stackrel{t \in [t_k, t_k+T_k]}{\Longrightarrow} \begin{bmatrix} x_{k+1} \\ y_{k+1} \\ z_{k+1} \end{bmatrix} = \begin{bmatrix} x_k + T_k u_k \\ y_k + T_k v_k \\ z_k + T_k (y_k u_k - x_k v_k) \end{bmatrix} $$
Inbetween steps:
$$\begin{cases} \dot{x} = u_k \\ \dot{y} = v_k \end{cases} \Longrightarrow \begin{cases} x(t) = x_k + (t - t_k) u_k \\ y(t) = y_k + (t - t_k) v_k \end{cases}$$
$$ \begin{aligned} \dot{z} & = (y_k + (t - t_k) v_k) \, u_k - (x_k + (t - t_k) u_k) v_k \\ & = (y_k \, u_k + (t - t_k) v_k \, u_k) - (x_k \, v_k + (t - t_k) u_k \, v_k) \\ & = y_k \, u_k - x_k \, v_k \end{aligned} $$ $$\therefore z(t) = z_k + (t-t_k) (y_k \, u_k - x_k \, v_k)$$
For that, I try to work backwards, i.e. $0 = (x_{k+1}, y_{k+1}, z_{k+1})$. From that, we find $(u_k, v_k)$ equal to $-\frac{1}{T} (x_k, y_k)$, which leads to $z_k = 0$. We can only go to the origin from an initial state $(x, y, 0)$.
So, how can we steer to origin from arbitrary point $(x_0, y_0, z_0)$ passing through to $(x, y, 0)$ with at least two intervals constant input values $(u_k, v_k)$?
You are pretty much there. You recognize that it is easy to drive the system to the origin from initial states where $z = 0.$ Your first goal is then to find a control action that drives the system to that plane. If $x_0 \neq 0$ or $y_0 \neq 0$, then the solution is clear. For instance, if $x_0 \neq 0$ set
$$u_0 = 0,\quad v_0 = \frac{z_0}{T_k\,x_0}$$
and you are assured that $z_1 = 0$. Clearly, we need to special case when $x_0 = y_0 = 0$, but that only requires one additional action prior to this which involves making at least one of these states non-zero. Just choose any non-zero action for $u$ and/or $v$. It is this exception that makes it at least two intervals of constant control action.
After this sample period, you can then do what you already suggested which is set
$$u_1 = -\frac{x_1}{T_1},\quad v_1 = -\frac{y_1}{T_1}.$$
Together that makes two intervals of control action with the exceptional case of three when the initial state is on the vertical axis $x_0 = y_0 = 0.$ If you are looking for a proof for why it must be at least two, you should search for a contradiction assuming this initial condition.