Let $X_{n} \in \mathbb{R}^{\infty}$ be a tight sequence of processes in metric space $(\mathbb{R}^{\infty}, l_{2})$ and for each $x\in\mathbb{Z}_{+}$ we have that $X_{n,x}\stackrel{d}{\to} Y_{x}$.
Does it follow the weak convergence of the process $X_{n}$ to $Y$?
Since the $\ell_2$ space is separable and complete, we know that we can extract convergent subsequences of a tight sequence. Therefore, it suffices to determine whether if $Y$ and $Y'$ are two $\ell_2$ valued random variables such that $Y_x=Y'_x$ in distribution, then $Y=Y'$ in distribution. But this is not true, even in $\mathbb R^2$, since two vectors can have the same marginals without sharing the same distribution.
However, what is true is the following: if $\left(X_n\right)_{n\geqslant 1}$ is a tight sequence in $\ell^2$ such that for any integer $d$, the vector $\left(X_{n,x}\right)_{x=1}^d$ converges to $\left(Y_{x}\right)_{x=1}^d$ in distribution as $n$ goes to infinity, then $X_n\to Y$ in distribution in $\ell^2$.
This is due to the fact that finite dimensional distribution characterize probability measures in $\ell_2$ since we can write an open ball as an intersection of sets depending only a finite number of coordinates. Indeed, we have $$B\left(x,r\right)=\bigcap_{N\geqslant 1}\bigcup_{m\geqslant 1}\left\{y\in\ell_2,\sum_{j=1}^N\left|x_j-y_j\right|^2 \lt r^2-\frac 1m\right\}.$$ Therefore, if two probability measures have the same finite dimensional distribution, they agree on finite unions/intersections of open ball. This proves that two probability distributions which have the same finite distributions agree on open balls.