The authors did not say anything like "$\mathbb R^{\mathbb N}$ is not high-dimensional enough." 

Rather, they said 

>"finite dimensional convergence [...] is too weak a result to develop genuinely high-dimensional inference methods". 

The meaning here is rather the opposite: $\mathbb R^{\mathbb N}$ is "too high" dimensional for the finite-dimensional convergence to work. 

Indeed, if one wants to study the behavior of $S_{n,p}:=S_n$ for large $n$ and $p$, then it is not enough to know the behavior of $S_{n,p}$ for large $n$ but only for a fixed finite set of values $p$. You already "understand that weak convergence (= convergence in distribution) in $\mathbb R^{\mathbb N}$ is equivalent to weak convergence of the finite dimensional marginals." So, the highlighted thesis follows. 

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On a somewhat positive note: In [this comment][1], the OP wrote: 

>The fact that finite dimensional convergence implies infinite dimensional convergence sounds like a nice feature in that regard (as it simplifies things by a lot). In the paper, and in your answer, this fact sounds like I huge disadvantage though.  

Of course, I said nothing of this sort. In fact, I did not talk about any advantages or disadvantages at all. 

What can actually be said on this matter is the following. The convergence of  the finite-dimensional distributions is of course **necessary** for the convergence of the distributions of the entire processes. Moreover, there are a number of results saying that, **with the additional tightness condition**, the convergence of the finite-dimensional distributions is **also sufficient** for the convergence of the distribution of the entire processes -- see e.g. [Theorems 7.1 and 13.1][2].

 


  [1]: https://mathoverflow.net/questions/481146/why-is-mathbb-r-mathbb-n-not-high-dimensional-enough/481149#comment1252850_481149
  [2]: http://cermics.enpc.fr/~monneau/Billingsley-2eme-edition.pdf