Q:

Convergence in topological space, coordinate-wise convergence

Let $(X_j,\tau _j), j\in I$ and $(Z:=\prod_{j\in I},\tau ^\times)$ be topological spaces.

Let, for some sequence $z_n\in Z, n\in\mathbb{N}$
  $$\forall j\in I\left (\pi _j(z_n)\xrightarrow[n\to\infty]{} \pi _j(z)\right )\quad (X_j,\tau _j) $$ where $z\in Z$. 
  Show that $z_n\xrightarrow[n\to\infty]{}z\quad (Z,\tau _j)$. $\pi _j:Z\to X_j, j\in I$ is the projection mapping.
  Show that the implication doesn't hold for $\tau ^\square$.

Essentially, if we let $z_n = (z_j^n)_{j\in I}, n\in\mathbb{N}$ and we have the given premise, is it true that coordinate-wise convergence is sufficient for convergence, at least for $\tau ^\times$?  
Define $x_n\to x$ in some topological space $(X,\tau)$ if for every open neighborhood (NH) $U\in\tau$ of $x$, there exists index $N\in\mathbb{N}$ s.t for every $n>N$, $x_n\in U$.  
By assumption, we have $\pi _j(z_n) = z_j^n\xrightarrow[n\to\infty]{}z_j=\pi _j(z)$. Letting $j\in I$, for every open NH $V_j\in\tau _j$ of $z_j$, exists index $N_j\in\mathbb{N}$ with $n>N_j$ implying $z_j^n\in V_j$.  
This means I can put together $z := (z_j)_{j\in I}$, but where would I use the fact that I'm operating w.r.t $\tau ^\times$?  
As a counter-example, why the implication doesn't hold for $\tau ^\square$, one is invited to look at $(Z := \prod_{j\in\mathbb{N}}\mathbb{R},\tau ^\square)$ with $z^n = (\underbrace{0,0,\ldots ,0}_{n-1}, n, n,\ldots),n\in\mathbb{N}$. from which it  seems that for every $j\in I$ $z_j^n\xrightarrow[n\to\infty]{}0$. Yet it's not sufficient for $z^n\xrightarrow[n\to\infty]{}0 =: z$?

Drafting space.  Assume $I$ infinite
Basis of product topology: $\mathfrak{B}^\times := \left\lbrace \prod_{j\in I}U_j : u_j\in\tau _j, U_j=X_j \mbox{ for almost all }j\in I\right\rbrace$.
Basis of box topology: $\mathfrak{B}^\square := \{\prod_{j\in I}U_j : U_j\in\tau _j\}$
Basis of neighborhood: $\mathfrak{B}_x := \{B\in\mathfrak{B} : x\in B\}$.  

A:

Indeed for the product topology $\tau^\times$ we have that coordinate-wise convergence is necessary and sufficient for convergence, while for the box topology $\tau^\square$ pointwise convergence is necessary (as the projections are continuous), but not sufficient. 
The example of the sequence $z^n \in Z= (\prod_{j \in \mathbb{N}} \mathbb{R}, \tau^\square)$, where $(z^n)_i = 0$ for $i < n$ and $(z^n)_i = n$ for $i > n$, is a fine example of this. For each fixed coordinate $m$, the sequence $(z^n)_m$ consists of $1,2,3,$ for the first $m$ terms, and $0$ after the $m$-th term. So every coordinate sequence is eventually $0$ (We can take $N= m$ for every neighbourhood of $0$ in that coordinate) and thus convergent to $0$.
But the neighbourhood $\prod_{j \in \mathbb{N}} (-1,1)_j$ in the box topology contains no $z^n$ at all, as $(z^n)_n =  n \notin (-1,1)$ for every $n$. So there is a neighbourhood of $0$ that does not contain a tail of the sequence so the sequence $(z^n)_n$ does not converge to the all $0$ sequence. 
If we have that $(z^n)_j$ converges to some $z_j$ for every $j \in J$, and we have the product topology, then $(z^n)$ does converge to $z := (z_j)_j$. We only have to consider basic open neighbourhoods for convergence so take any $U = \prod_j U_j$ where every $U_j$ is open in $X_j$ containing $z_j$, and  we have a finite subset $I \subseteq J$ such that $U_j = X_j$ for $j \notin I$. This means we only hav finitely many constrained coordinates that we have to get the tails of the $x^n$ into:
For every $i \in I$ we pick $N_i$ such that for all $n > N_i$ we know that $(z^n)_i \in U_i$ by convergence in  this coordinate $i$.
Now take $N = \max(N_i, i \in I)$ which is a maximum of finitely many $N_i$, so finite as well. And if $n >N$, consider any coordinate $j$. If $j \in I$ then  $(z^n)_j \in U_j$ as $n > N \geq N_j$. If $j \notin I$, $(z^n)_j \in U_j = X_j$ is trivial (no constraint by the form of the type of neighbourhood). 
In all cases $(z^n)_j \in U_j$, so $z^n \in U$ for $n > N$ as required. This shows convergence of the $z^n$ to $z$.

