According to https://mathoverflow.net/a/8994/4600,
$\log_2$ of the number of topologies on $n$ elements is $\sim n^2/4$.
So let's say there are $2^{n^2/4}$ topologies.
Then the number of topologies on $X_1=[n]=\{0,\dots,n-1\}$ and $X_2=[n]$ is
$$
2^{n^2/4}\cdot 2^{n^2/4} = 2^{n^2/2}.
$$
Consider now sets $R$ which contain $(k,0)$ for each $k<n$. There are
$$
2^{n(n-1)}
$$
such sets. But there are only
$$n^n = 2^{n\log_2 n}$$
functions, so the total number of $(\tau_1,\tau_2,f)$ is only
$$
2^{n^2/2}2^{n\log_2 n} < 2^{n(n-1)}
$$
for large $n$.
So there are just too many sets $R$.

**Edit:**
This is rather complicated compared to Adam Przeździecki's solution, but it does have the virtue of showing that "**closure of**" could be replaced by any other method of uniquely obtaining $R$ from $f$, $\tau_1$, and $\tau_2$.