How do you identify a world line is in proper time? I’m attempting to answer the question:

Show that the world line:
$$ x(\lambda) = \begin{pmatrix}c\lambda \\ 0 \\ 0 \\ c\sin(\omega \lambda)\end{pmatrix}$$
where $ (\lambda,\omega>0) $
isn’t written in proper time.

The claim in the solution is that, if $\lambda $ was proper time, then we would have:
$$\langle\frac{dx(\lambda)}{d\lambda},\frac{dx(\lambda)}{d\lambda}\rangle_{M} = -c^2,$$
I can evaluate this result and show that it doesn’t hold for the values bounded as they are.
I don’t understand where this result has come from though and can find no reference to anything resembling it anywhere.
If anyone can give any insight into where this result has come from and/or if it is generally the case, it would be really helpful. I have been really stuck on this for a while.
 A: The Wikipedia page has a section in which it says

If $t, x, y, z$, are parameterised by a parameter $\lambda$, [the proper time interval] can be written as
\begin{equation}
\Delta \tau =\int {\sqrt {\left({\frac {dt}{d\lambda }}\right)^{2}-{\frac {1}{c^{2}}}\left[\left({\frac {dx}{d\lambda }}\right)^{2}+\left({\frac {dy}{d\lambda }}\right)^{2}+\left({\frac {dz}{d\lambda }}\right)^{2}\right]}}\,d\lambda
\end{equation}

This is essentially the definition of proper time $\tau$ — except that you're always allowed to shift the point at which $\tau=0$.  This is just a time offset that has no physical significance, and is related to the fact that we're not being too explicit about the limits of the integral.
Now, when your question asks whether the worldline is "written in proper time", basically that will be true if $\lambda$ is precisely $\tau$ (at least, up to a time offset).  Looking at that integral, you can see that this is the case whenever the integrand is precisely 1.  That is, $\lambda$ is precisely $\tau$ (up to a time offset) if and only if
\begin{equation}
\left({\frac {dt}{d\lambda }}\right)^{2}-{\frac {1}{c^{2}}}\left[\left({\frac {dx}{d\lambda }}\right)^{2}+\left({\frac {dy}{d\lambda }}\right)^{2}+\left({\frac {dz}{d\lambda }}\right)^{2}\right] = 1.
\end{equation}
Multiply both sides of this equation by $-c^2$, and you get exactly the condition in the question above:
\begin{equation}
\left\langle\frac{dx(\lambda)}{d\lambda},\frac{dx(\lambda)}{d\lambda}\right\rangle_{M} = -c^2.
\end{equation}
So yes, this condition is very generally what you need to check to see if $\lambda$ is the proper time of a worldline.
A: As you can easily see, the world line would parametrized by proper time if you were to drop the $z$-component. Which means that $\lambda$ would be the proper time if the observer on the given world line would not be moving (w.r.t. the chosen coordinate system).
But since said observer is moving (oscillating) there will be a time dilation effect between the resting and the moving observers.
The situation here of course gets complicated by the fact that the resting observer and the oscillating observer are not related by a simple Lorentz boost, since the oscillating observer is undergoing acceleration. There is however of course also a proper time which this oscillating observer is experiencing and it is given by the Minkowski-product $\langle\frac{dx(\lambda)}{d\lambda},\frac{dx(\lambda)}{d\lambda}\rangle_{M}$ that you calculated.
Does that help you?
