Let's say a particle moves in plane according to $\tag{1} r = f(t), \theta=g(t)$
Let $A(t)$ be the area of the region swept out by the position vector from a fixed time point (e.g. 0), to a later time $t$. We want to prove that $\tag{2}A'(t) = \frac{1}{2}r^2\frac{d\theta}{dt}$
Tom Apostol proceeds by assuming that it is possible to eliminate $t$ from equations in $1$, and express $r$ as a function of $\theta$, i.e. $\tag{3} r = R(\theta)$ Then he proceeds to prove the $2$.
However, why is it justified to assume that such $R$ exists?
I can imagine a case where a particle is moving between $[\theta_1, \theta_2]$ (back and forth), where $r$ is increasing with time, in which case there is no such $R$, as far as I can see.
Am I missing something?
Your intuition is correct, the existence of such a function $R$ isn't guaranteed. Indeed, one has $r = f(t) = f(g^{-1}(\theta))$, hence $R = f \circ g^{-1}$, but $g$ might be non-invertible. A simple example is given by the straight line parametrized by $$ \begin{cases} r = t \\ \theta = 1 \end{cases} $$ The case $\theta = const$ as above is actually quite special and can be circumvented by arguing that $\dot{\theta} = 0$, whence $A = const$. The other cases correspond to non-injective $g$ such as $\theta = \sin t$ for instance, but again this problem may be bypassed by cutting the domain of integration in such a way that $g$ is a local/piece-wise bijection.
In the end, the origin of the present obstacle is nothing else than a mathematical artifact due to algebraic manipulations in the change of variable $t \to \theta$ when computing $$ A(t) = \int \frac{1}{2}r^2\dot{\theta} \,\mathrm{d}t = \int \frac{1}{2}r^2 \,\mathrm{d}\theta, $$ while the original integrand, namely $A'(t) = \frac{1}{2}f(t)^2g'(t)$ is perfectly defined.