= Solution
Choose the <potential right-censoring times> to include every observed <right censoring> time up to $t$, and also the endpoints 0 and $t$. Then no individual is censored strictly inside $(c_{k-1},c_k)$. Use an explicit endpoint convention: let $r_k$ count individuals still event-free after any events at $c_k$, but before censoring at $c_k$. If $d_k$ events occur in $(c_{k-1},c_k]$, the number at the beginning of that interval, after censoring at its left endpoint, is $r_k+d_k$.
The no-censoring interval estimate from the preceding parts is therefore $r_k/(r_k+d_k)$. The possible event mass at zero contributes $r_0/(r_0+d_0)$. Multiplying yields
$$
\boxed{\widetilde F(t)=\frac{r_0}{r_0+d_0}\prod_{k=1}^h\frac{r_k}{r_k+d_k}.}
$$
For a continuous event-time distribution, normally $d_0=0$. Within interval $k$, put $b_k=r_k+d_k$ and let its successive event counts be $e_1,\ldots,e_m$. Its <Kaplan–Meier estimator> factors telescope:
$$
\prod_{j=1}^m\left(1-\frac{e_j}{b_k-\sum_{\ell<j}e_\ell}\right)
=\frac{b_k-\sum_j e_j}{b_k}=\frac{r_k}{r_k+d_k}.
$$
The same calculation at zero proves
$$
\boxed{\widetilde F(t)=\widehat F(t).}
$$
The paper's phrase “at risk at $c_k$” does not specify its side of an event at that endpoint. If it denotes the conventional pre-event <risk set> size $r_k^-$ instead, and $e_k$ events occur exactly at $c_k$, substitute $r_k=r_k^--e_k$. The interval factor is then $(r_k^--e_k)/(r_k^-+d_k-e_k)$; at zero it is $(r_0^--d_0)/r_0^-$. This keeps the events-before-censoring convention consistent. After an empty <risk set>, stop the product; if the fitted survival has reached zero, retain zero rather than creating a $0/0$ factor.
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