It seems quite "obvious" that if you sort a $n$-tuple of i.i.d. variables (assuming they have an order and ties do not exist - which is true with probability $1$ for, say, $U(0, 1)$) that the resulting permutation is a uniformly random one - after all what would be the source of any bias? But obvious does not mean proven, and I've struggled for a bit to come up with a proof. I think I have done so now, but I feel like it is overly complicated, a bit informal, and most importantly, "backwards". Is there a simpler proof?
Let $r$ be a tuple of $n$ i.i.d. random variables (with an order and no equal elements with probability $1$). Let $\sigma$ be a permutation chosen at uniformly at random from the $n!$ possible permutations on $n$ elements, independent from $r$.
Note that $\sigma(r)$ samples from the same distribution as $r$, since any blind (independent of $r$) reordering of i.i.d. variables is indistinguishable from the original. We can thus see that $s = \sigma(\mathrm{sort}(r))$ also samples from the same distribution as $r$, since $\sigma$ is a random permutation which destroys all information about order prior to the permutation, making the sort irrelevant and the two indistinguishable.
By sorting $s$ we get $\sigma^{-1}$, the inverse of a randomly chosen permutation, assuming there are no duplicate elements. But since there is a one-to-one mapping between $n$-permutations and their inverses, and $\sigma$ was chosen uniformly at random, $\sigma^{-1}$ samples from the same distribution as $\sigma$.
Since $s$ is distributed like $r$, and $\sigma^{-1}$ is distributed like $\sigma$ we can thus conclude that by sampling $r$ and sorting it we can sample from a $n$-permutation uniformly at random.
First you need to ensure that ties are not possible. If $P(X_1 = x) = 0$ for all $x$, then this is satisfied.
Thus the order of $X = (X_1, \dots, X_n)$ defines a well defined random permutation $s(X)$ meaning that $X_{s(X)(1)} < \dots < X_{s(X)(n)}$. Now we want to show that $s(X)$ has the uniform distribution on $S_n$, i.e. that for every $\sigma \in S_n$, $P(s(X) = \sigma) = \frac{1}{n!}$. To show this, it suffices to show that $P(s(X) = \sigma) = P(s(X) = id)$ for all $\sigma\in S_n$. Note that $\{s(X) = \sigma\} = \{s(X_{\sigma(1)}, \dots, X_{\sigma(n)}) = id\}$. So it suffices to show that $(X_1, \dots, X_n)$ and $(X_{\sigma(1)}, \dots, X_{\sigma(n)})$ have the same distribution. But this is true by the $\pi-\lambda$ theorem since by the i.i.d assumption, their joint laws agree on sets of the form $A_1 \times \dots \times A_n$ with $A_1, \dots, A_n \subset \mathbb{R}$.