How do we prove that the 4-acceleration transforms as a 4-vector in Special Relativity? In order to define the acceleration of a body in its own frame, we need to first prove that the acceleration is a four-vector so that its dot product with itself can then be labeled as acceleration squared in the rest frame. For velocity and displacement vectors, we can show that they have a constant dot product. But how do we prove that for acceleration?
 A: Is it not so by definition?
$$
{\bf a}= \frac {d{\bf v}}{d\tau}
$$
where
$$
{\bf v}= \frac{d{\bf x}}{d \tau}
$$
is a 4-vector and $\tau$ is a scalar.
A: Since you accept that four-velocity is a four-vector, this is an argument that four-acceleration is a four-vector:
$$a^{\mu}=\lim _{h\rightarrow 0} \frac{v^{\mu}(\tau+h)-v^{\mu}(\tau)}{h}$$.
The path is parametrized by $\tau$, the proper time, which is a scalar because it's equal to the spacetime interval (upto maybe a sign)
You can imagine carrying out this limit calculation in two different frames. If you're doing a numerical calculation, you will take $h$ to a small finite number.
The numerator will be a difference of four-vectors, hence it is a four-vector. The denominator is a scalar. Hence, the fraction is a four-vector.
This isn't a proof of course. A proof will try to argue that the limit of the sequence four-vectors, as $h\rightarrow 0$, will also be a four-vector (which makes sense sort-of)
EDIT Ok so, let's say $v^{\mu}(h)$ is a sequence of four-vectors paramerized by a real parameter $h$. Let $v^{\nu} (h)=\Lambda v^{\mu}(h)$, $\Lambda$ is a Lorentz transform.
Then,
$$\lim_{h\rightarrow 0}v^{\nu} (h)= \lim_{h\rightarrow 0}\Lambda v^{\mu}(h)$$
$$=\Lambda \lim_{h\rightarrow 0} v^{\mu}(h)$$
So you see, the limit in one frame is the Lorentz transform of the limit in another frame (We could pull $\Lambda$ out of the limit because it's a constant matrix)
A: In physics, we prove things with experiments. Four-vectors are components of a mathematical model. Does that model pass experimental test? Yes, in many cases. We use it because it passes those tests, not because of any mathematical proof.
