Let's call $\vec{OM}(t) = \mathbf{r}(t)$. 

**Proof of Point 1.** From 5. you know that $|\mathbf{r}(t)|$ is constant and thus the derivative of its square is equal to zero

$0 = \dfrac{1}{2} \dfrac{d |\mathbf{r}|^2}{dt} = \dfrac{1}{2} \dfrac{d }{dt}(\mathbf{r}\cdot\mathbf{r}) = \dfrac{d \mathbf{r}}{dt} \cdot \mathbf{r} = \mathbf{v} \cdot \mathbf{r}$,

and thus $\mathbf{v}$ is orthogonal to $\mathbf{r}$.

**Proof of Point 2.** This do not depend on the peculiar motion. For every motion, $\mathbf{r}(t)$, we can decompose velocity and acceleration in a local unit vector basis, the Fresnel basis, formed by the tangent vector $\mathbf{\hat{t}}$, the normal vector $\mathbf{\hat{n}}$, and the binormal vector $\mathbf{\hat{b}}$.

Velocity is a vector always pointing in the direction tangent to the trajectory, $\mathbf{v}(t) = v(t) \, \mathbf{\hat{t}}(t)$. Its time derivative is the acceleration that reads

$\mathbf{a} = \dfrac{d \mathbf{v}}{dt} = \dfrac{d }{dt}( v \mathbf{\hat{t}} ) = \dfrac{d v}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt}
$

where $\dfrac{d v}{d t}$ is the magnitude of the tangential acceleration, and the time derivative of the tangent vector is orthogonal to the tangent vector itself (since $0 = \frac{d}{dt}(\mathbf{t} \cdot \mathbf{t} ) = 2 \dfrac{d \mathbf{\hat{t}}}{dt} \cdot \mathbf{\hat{t}} )$ and can be expressed as

$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}}$,

where $k = \dfrac{1}{R}$ is the curvature of the trajectory, $R$ is the local radius of curvature, $\mathbf{\hat{n}}$ is the unit normal vector pointing towards the center of circle of curvature. Thus, acceleration can be written as

$\mathbf{a} = \dfrac{d v}{dt} \mathbf{\hat{t}} + k \, v^2 \mathbf{\hat{n}} = \dfrac{d v}{dt} \mathbf{\hat{t}} + \dfrac{ v^2 }{R} \mathbf{\hat{n}}$.

**What is still missing here.** In the derivation above, I made no use of the information that the motion must be planar, i.e. orthogonal with a desired constant vector $\mathbf{\hat{k}}$ (so the motion could occur here on the surface of a sphere), and some details to prove the fact that the radius of curvature and the curvature is constant.