Rudin in PMA, gives proof of chain rule as this:
Theorem: Suppose $f$ is continuous on $[a,b]$, $f'(x)$ exists at some point $x\in [a,b]$, $g$ is defined on an interval $I$ which contains the range of $f$, and $g$ is differentiable at the point $f(x)$. If $$\begin{align} h(t)=g(f(t)) &&(a\le t\le b),\end {align}$$ then $h$ is differentiable at $x$, and $h'(x)=g'(f(x))f'(x)$.
Proof: Let $y=f(x)$. By the definition of derivative, we have $$f(t)-f(x)=(t-x)[f'(x)+u(t)],\tag4$$$$g(s)-g(y)=(s-y)[g'(y)+v(s)],\tag5$$ where $t\in [a,b], s\in I$, and $u(t)\to0$ as $t\to x$, $v(s)\to0$ as $s\to y$. Let $s=f(t)$. Using first (5) and then (4), we obtain $$\begin{align}h(t)-h(x)&=g(f(t))-g(f(x))\\&=[f(t)-f(x)]\cdot[g'(y)+v(s)]\\&=(t-x)\cdot[f'(x)+u(t)]\cdot[g'(y)+v(s)],\end{align}$$ or if $t\ne x,$ $$\dfrac{h(t)-h(x)}{t-x}=[g'(y)+v(s)]\cdot[f'(x)+u(t)].\tag 6$$ Letting $t\to x$, we see that $s\to y$, by the continuity of $f$, so that the right side of $(6)$ tends to $g'(y)f'(x)$.
But, these errata, on page 5, under heading P.105, say that "Rudin’s proof skirts the point that makes proving the Chain Rule difficult: that $f(t)$ may take on the value $f(x)$ infinitely often in the neighborhood of $x$". My question is, does this issue remain even after errors as mentioned in link are taken care of?
Also, does Apostol's proof of chain rule take care of possibility of infinitely many $f(t)=f(x)$ in neighborhood of $x$?
There is no issue in the proof. Once $x$ and $y$ are fixed, the functions $u$ and $v$ exist uniquely and are continuous with zero values at $x$ and $y$. Perhaps in some other proofs obtaining the same value infinitely often is an issue, but no in this one.
Apostol's proof is fine as well. The proofs are valid even if $f$ is a constant.