The coordinates of an event in spacetime are given by the 4-vector $(ct, \mathbf{r})$, where $\mathbf{r}$ is the spacial coordinates of the event. This 4-vector can be seen as 4-displacement of a worldline from the defined origin of the reference frame we're in at time $t$.
It seems sensible that $\frac{d}{dt}(ct,\mathbf{r})$ should give 4-velocity of the worldline, but instead everything I've read has stated that we differentiate with respect to the worldline's proper time $\tau$ instead, and yet I so far haven't seen any explanation as to why. This answer here on the Stack Exchange simply says we do it because it maintains the Lorentz invariant. However, why would proper time be invariant under the Lorentz transformation and other times wouldn't?
Consider $\mathbf{x^\mu}=(ct,x,y,z)^T$, which I differentiate with respect to time $t$ to get $\mathbf{v}=(c,v_x,v_y,v_z).$ Let's check if this is Lorentz invariant:
$$ \begin{bmatrix} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} c\\ u_x\\ u_y\\ u_z \end{bmatrix} =\begin{bmatrix} c\gamma-\beta\gamma u_x\\ -\beta\gamma c+\gamma u_x\\ u_y\\ u_z \end{bmatrix} =\mathbf{v'} $$
$$ (c\gamma-\beta\gamma u_x)^2-(-\beta\gamma c+\gamma u_x)^2=c^2\gamma^2+\beta^2\gamma^2u_x^2+\beta^2\gamma^2c^2+\gamma^2u_x^2 =c^2\gamma^2(1+\beta^2)-u_x^2\gamma^2(1+\beta^2)=c^2-u_x^2 \\ \therefore \mathbf{v'}\cdot\mathbf{v'}=c^2-u_x^2-u_y^2-u_z^2=\mathbf{v}\cdot\mathbf{v} $$
Therefore, $\mathbf{v}$ is Lorentz invariant. Why then, do we reject it as the velocity 4-vector?