Does the Glivenko-Cantelli theorem work back and forth? If we have a sample $X_1, X_2, \ldots, X_n \sim F$ then $\hspace{1mm}sup_x|F_n(x) -F(x)|\xrightarrow{a.s./p}0$.
Now, if I can come up with a theoretical cdf $F$ such that $\hspace{1mm}sup_x|F_n(x) -F(x)|\xrightarrow{a.s./p}0\hspace{1mm}$ am I allowed to claim that $X_1, X_2, \ldots, X_n \sim F$ ?
If not, could you give me a counterexample?
 A: It depends on what you mean.  If you mean that $X_1,\dots,X_n$ are iid from some distribution $G$, and it turns out that $\sup_x |F_n(x)-F(x)|\stackrel{p}{\to}0$, then yes, you can conclude that $F=G$.
Proof: The Glivenko-Cantelli theorem says $\sup_x |F_n(x)-G(x)|\stackrel{p}{\to}0$, so by the triangle inequality $\sup_x |F(x)-G(x)|\stackrel{p}{\to}0$ and as the LHS doesn't depend on $n$, $F=G$ (on a set of probability 1).
On the other hand, it's quite possible to have a non-iid random mechanism that generates $X_i$ such that $\sup_x |F_n(x)-F(x)|\stackrel{p}{\to}0$
For example, let $M$ be a real-valued random variable with $E[M]=0$, and let $X_i = M+\epsilon_i$, where $\epsilon_i$ are iid mean zero random variables. Conditional on $M=m$, $X_i$ are iid with the same CDF as $\epsilon_i+m$ and the Glivenko-Cantelli theorem says that $F_n$ converges uniformly to the CDF of $\epsilon_i+m$.  This isn't the CDF of any individual $X_i$ (which would be that of $\epsilon_i+M$).
That's the set-up of de Finetti's theorem and the Hewitt-Savage 0-1 law: an exchangeable sequence of random variables looks iid, but the limit is random rather than fixed.
There are other ways this can go wrong. Suppose $X_i$ are identically distributed with cdf $F$ but not independent. It's possible (under suitable conditions on long-range dependence) that the empirical cdf $F_n$ converges uniformly to $F$. It would still not be true that $X_i$ are iid from $F$.
Or, the distribution of $X_i$ could vary periodically. As $n\to\infty$, the CDF $F_n$ would still converge uniformly to a CDF of the mixture of $X_i$ over a period.  For example, if $X_n\sim N(1,1)$ for even $n$ and $N(0,1)$ for odd $n$, $F_n$ will converge to the CDF of a 50:50 mixture of $N(0,1)$ and $N(1,1)$.
