Let AB be a segment of non-zero length in K, parallel to the x-axis, with xA < xB. We saw before that for the corresponding points A' and B' in K', respectively, marked out at any t in K, x'A' < x'B' holds.
In K', let's send a light signal from A' toward B'. It will arrive at B' after some time Δt' > 0. For the corresponding duration in K, Δt > 0 must hold due to causality. (Relative to K', the light signal meets first A' and then B', so it happens in the same order relative to K too.)
Since the light signal is moving at a constant velocity relative to K', it has to do likewise relative to K as well. Furthermore, both A' and B' are moving at v relative to K, and when the light signal is emitted, B' is ahead of A'. Later the light signal meets B', which then means it is traveling in K along a straight line parallel to the x-axis, in the positive direction. And by law, at the constant speed c. The fact that it does catch up with B' implies:
(Theorem) v < c.
The relationship between inertial frames and rigid objects entails that particles too have exactly the same speed limit.