= Solution
Choose the two <inertial frames> to have coincident origins at $t=t'=0$, and let the origin of $S'$ have $x=vt$. The <Lorentz transformation> is
$$
\boxed{x'=\gamma_v(x-vt),\qquad
t'=\gamma_v\left(t-\frac{vx}{c^2}\right),\qquad
\gamma_v=\frac1{\sqrt{1-v^2/c^2}}.}
$$
Here $\gamma_v$ is the <Lorentz factor>. Along the particle's trajectory, $dx/dt=u$, so
$$
\frac{dx'}{dt}=\gamma_v(u-v),\qquad
\frac{dt'}{dt}=\gamma_v\left(1-\frac{uv}{c^2}\right)>0.
$$
Dividing gives the <relativistic velocity-addition formula>
$$
\boxed{u'=\frac{u-v}{1-uv/c^2}.}
$$
The denominator is positive because $|uv|<c^2$. Direct algebra yields
$$
1-\frac{u'^2}{c^2}
=\frac{(1-u^2/c^2)(1-v^2/c^2)}{(1-uv/c^2)^2}>0,
$$
since both factors in the numerator are positive. Consequently $\boxed{|u'|<c}$: a subluminal velocity remains subluminal in every such <inertial frame>.
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