Assertion carries no weight.
Learn a little. This is called 'proof'. No assertion here.
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Eq. 1) Einstein defines x' = x-vt. (not me, Einstein. see section 3 of
Electrodynamics)

Eq. 2) Einstein takes x' to be infinitessimally small. (not me, Einstein)

The Lorentz transform, as given in  Electrodynamics, is
Eq. 3a) xi = (x-vt) / sqrt(1-v^2/c^2) and
Eq. 3b) tau = (t-vx/c^2(/sqrt(1-v^2/c^2)

which you cannot sensibly deny, you've used it.

So far,  I have done nothing.
Now.
Any child learning algebra will agree that if
x' = x-vt (eq. 1)
then
x = x' +vt. (eq. 4)
If that child does not agree, that child is not pass his/her examinations.

When x' is infinitessimally small ( eq 2),
x' = 0+h as h tends  to zero.
Eq. 4) x = h + vt.

If we substitute x for its value in (3a)

Eq. 5)
xi =  (h+ vt-vt) / sqrt(1-v^2/c^2)
   = h / sqrt(1-v^2/c^2)
   = 0 in the limit of h.

So, in time t, the displacement of xi from the origin of the 'stationary'
(inertial, non accelerative, fancy named)  frame is nearly nothing = 0,
hence

Eq. 6) v = 0 for all v, t > 0.

Applying our value we have found for v (eq. 6) in (eq. 3b),

Eq. 7) tau = (t - 0.x/c^2)/sqrt(1- 0^2/c^2)  = t.

Conclusion:
The Lorentz transforms are valid for all v  = 0, otherwise they are
nonsense.

Quad Erat Demonstrandum.
You are invited to fault the logic. You are not invited to call me moron.