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 Frenet 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.
EDIT
Let's introduce the constraint of planar motion. Motion occurs in a plane passing through the origin and perpendicular to the constant vector $\mathbf{\hat{k}}$.
So velocity itself is perpendicular to $\mathbf{\hat{k}}$ as well, and so the tangent vector $\mathbf{\hat{t}}$ and its derivative $d\mathbf{\hat{t}} / dt$.
For planar motions, its convenient to introduce polar coordinates, so that you can write
$\mathbf{r} = r \mathbf{\hat{r}}$
$\mathbf{v} = \dot{r} \mathbf{\hat{r}} + r \dot{\theta} \mathbf{\hat{\theta}}$
$\mathbf{a} = \left[ \ddot{r} - r\dot{\theta}^2 \right] \mathbf{\hat{r}} + \left[ 2 \dot{r} \dot{\theta} + r \ddot{\theta} \right] \mathbf{\hat{\theta}}$ .
Now, let's introduce the assumption that $r = R = \text{const.}$ to get
$\mathbf{v} = R \dot{\theta} \mathbf{\hat{\theta}} = R \dot{\theta} \mathbf{\hat{t}} = v \mathbf{\hat{t}}$
$\mathbf{a} = - R \dot{\theta}^2 \mathbf{\hat{r}} + R \ddot{\theta} \mathbf{\hat{\theta}} = - R \dot{\theta}^2 \mathbf{\hat{n}} + R \ddot{\theta} \mathbf{\hat{t}} = \dfrac{v^2}{R} \mathbf{\hat{n}} + a_t \mathbf{\hat{t}} = k \, v^2 \mathbf{\hat{n}} + a_t \mathbf{\hat{t}}$,
and thus the unit normal vector points towards the center of the circle $\mathbf{\hat{n}} = -\mathbf{\hat{r}}$, the curvature $k = \frac{1}{R}$ is constant, and $a_t = R\ddot{\theta}$ is the tangent acceleration.
Comment. You ask to prove it without using coordinates, but you don't be afraid of using coordinates. I agree with you that
- physics, as well as geometry, is coordinate independent;
- physical laws must be written in abstract notation (vector/tensor equations), because it's the most general form, and doesn't hide the tensor form of the equations writing only the coordinates;
but remember that you can introduce coordinates if it useful to do so, i.e. it reduces the amount and the complexity of the computations. Remember also that if you prove something using a set of coordinates, this is true in general.
