Prove that a non-zero acceleration is perpendicular to a constant speed

You can apply the product rule when differentiating $\vec v\cdot\vec v =$constant.

The intuitive idea is that $\vec v(t)$ traces out a curve on a sphere centered at the origin (i.e., picture $\vec v(t)$ as a moving radius vector), while $\vec v'(t)=\vec a(t)$ is tangent to the sphere, and hence perpendicular to the radius at the point of tangency, namely $\vec v(t)$.

There is also a geometric description of what this says in terms of the original curve, say $\vec x(t)$, of which $\vec v(t)$ gives the velocity. Since $\vec v(t)$ gives a vector tangent to the trajectory of $\vec x(t)$, and in your case $\vec a(t)$ is perpendicular to $\vec v(t)$, $\vec a(t)$ is perpendicular to the trajectory of the original curve $\vec x(t)$.

Acceleration can occur for 2 reasons. In general, the component of $\vec a(t)$ in the direction of the trajectory (or opposite this direction) tells you how the speed is changing, while the component of $\vec a(t)$ perpendicular to the trajectory tells you how the direction is changing. So, again, in case the speed is constant, $\vec a(t)$ is perpendicular to the curve. Its direction tells you which direction $\vec x(t)$ is turning, while its magnitude tells you how quickly $\vec x(t)$ is changing direction. This last quantity is a constant multiple of the curvature of $\vec x(t)$.

The shortest way is to notice that:

$$v^2=\vec{v}\cdot\vec{v}$$

After differentiating: $$2v\frac{dv}{dt}=2\vec{v}\cdot\frac{d\vec{v}}{dt}$$ Done.

If $v(t) = \bigl(x_1(t),x_2(t),\ldots,x_n(t)\bigr)$ (I don't know what dimension you are working on), then the speed is given by $$||v(t)|| = \sqrt{ x_1(t)^2 + \cdots + x_n(t)^2}.$$ Since this is constant, say equal to $k$, then you have $$k^2 = x_1(t)^2 + \cdots + x_n(t)^2.$$

Now take derivatives with respect to $t$. Compare that with $v\cdot a$ (remembering that $a(t) = v'(t)$).

(Of course, if $\frac{dv}{dt}$ is zero, then it is also perpendicular to $v(t)$, since $\mathbf{0}$ is perpendicular to every vector...)