= Solution
Take $S'$ to move along $+x$ at velocity $V$, with the origins coincident at zero time. The <Lorentz transformation> is
$$
x'=\gamma_V(x-Vt),\qquad t'=\gamma_V(t-Vx/c^2),\qquad
\gamma_V=(1-V^2/c^2)^{-1/2}.
$$
Differentiating along the particle trajectory gives the <relativistic velocity-addition formula>
$$
\boxed{u'=\frac{u-V}{1-uV/c^2}.}
$$
For a timelike particle, $|u|<c$ and $|V|<c$. Direct algebra gives
$$
1-\frac{u'^2}{c^2}
=\frac{(1-u^2/c^2)(1-V^2/c^2)}{(1-uV/c^2)^2}>0,
$$
so \b[$|u'|<c$], in particular $u'<c$. Strictness requires a massive, subluminal particle; for a photon with $u=\pm c$, the transformation preserves $u'=\pm c$ instead.
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