Note: The term equilibrium is different from steady-state w.r.t chemical reactions. Steady state is the right term for the above example. Equilibrium is used in the sense of forward and reverse reactions in a single reversible reaction. The above example considers two irreversible reactions — production and degradation.
... ... the time required for the molecule to shift halfway from its old to its new equilibrium concentration is equal to its normal half-life
This is not true if the degradation reaction is zeroth order or if these reactions are nonlinear. But for a linear production degradation set of reactions ($X$ is a biomolecule formed at a constant rate $\alpha$ and degraded at a first order rate $\beta$ ), the reaction rate can be represented by the following equation.
$$\frac{d[X]}{dt}\bigg|_1=\alpha _1 - \beta [X]$$
At steady state $\frac{d[X]}{dt}=0$, which means $[X_{ss_1}]=\Large\frac{\alpha _1}{\beta}$
Half life as defined, by setting $\alpha$=0 and solving the differential equation = ${\Large \frac{log(2)}{\beta}}$
When the formation rate is changed to a new value:
$$\frac{d[X]}{dt}\bigg|_2=\alpha _2 - \beta [X]$$
and $[X_{ss_2}]=\Large\frac{\alpha _2}{\beta}$
From the above equation following equation can be deduced (by integrating the differential equation):
$$-\frac{1}{\beta}.log(\alpha _2 - \beta [X]){\large|}_{X_{initial}} ^{X_{final}}= t_{\frac{1}{2}} $$
when you substitute $X_{initial}=X_{ss_1}=\frac{\alpha _1}{\beta}$ and $X_{final}=\Large\frac{\frac{\alpha _2}{\beta} - \frac{\alpha _1}{\beta}}{2}$ (halfway between new steady state and old steady state)
you would obtain: $$t_{\frac{1}{2}}= -\frac{1}{\beta}log\left(\frac{\alpha _2 -\beta\frac{\alpha _1}{\beta}}{\alpha _2 - \beta \frac{\frac{\alpha _2}{\beta} - \frac{\alpha _1}{\beta}}{2}}\right) = \frac{log(2)}{\beta} $$
