Net convergence over product and box topology Let $(x_\alpha)_{\alpha\in \Lambda}$ a net in the topological space $X = \prod_{i\in I}X_i$.
Is it true that $x_\alpha \rightarrow x$ iff $p_i(x_\alpha) \rightarrow p_i(x)$ for every $i \in I$ with the product topology and box topology?
I think this is true for the product topology, but not for the box topology.
First case:
An open set $U$ of $X$, is $U=\prod_{i\in I} U_i$, where only for finite $i\in I$, $U_i \neq X_i$, so if I take an open set $U_j$ for a $j \in I$, such that $p_j(x) \in U_j$ then $x \in U$ with $U = \prod_{i \in I}U_i$ with $U_i=X_i$ if $i \neq j$, which is open. Since $x_\alpha \rightarrow x$, then it exists $\alpha_0\in \Lambda$ such that $x_\alpha \in U$ for every $\alpha > \alpha_0$. In particular, for every $\alpha > \alpha_0$, $p_j(x_\alpha) \in U_i$ so $p_j(x_\alpha) \rightarrow p_j(x)$.
Likewise, if $p_j(x_\alpha) \rightarrow p_j(x)$ for every $j \in I$, given an open set $U \subset X$, $U=\prod_{i\in I} U_i$, where only for finite $i\in I$, $U_i \neq X_i$ such that $x\in U$. Let's say that this final set is $A=\{i\in I/ U_i\neq X_i\}$. So, for every $i\in A$, exists $\alpha_i \in \Lambda$ such that $\alpha > \alpha_i \Rightarrow p_i(x_\alpha) \in U$. So if I take $\alpha_0= \max(\alpha_i)$, it's clear that $x_\alpha \in U$ for every $\alpha > \alpha_0$.
Second case:
I think the first part is pretty much the same (and I think it's also a good argument to say that this holds true since the projections are continuous), but in the second case I can't take maximum since there might be infinite $\alpha_i$, but I can't build a counter example.
 A: As a general consideration: Net convergence uniquely determines the topology (in contrast to sequence convergence which need not). And the product topology is characterised (some would say "defined", even) by the fact that net convergence is equivalent to convergence in every coordinate space. So the box topology, which is generally a lot stronger, cannot have that same characterisation.
In the countable product case we can already see this with sequences (general nets not needed): 
Let $\Box_{i=0}^\infty \Bbb R_i$ be the box product of countably many copies of $\Bbb R$ and $\Bbb R^{\Bbb N}$ the same set, but in the product topology.
Then $x_1 = (1,1,1,1,\ldots)$, $x_2=(0,1,1,\ldots,)$, $x_3=(0,0,1,1,1\ldots)$ and generally, $x_n$ is the point/sequence that is $0$ on coordinates $0$ to $n-1$, and $1$ onwards; we see that $(x_n) \to 0$ (the all-$0$ point/sequence) in the product topology (as in each coordinate $m$ the sequence $(\pi_m[(x_n)])_n$ is initially $1$'s and then all $0$'s after the $m$-th term, so converges to $0$. But the box neighbourhood $\Box_{i=0}^\infty (-\frac12,\frac12)_i$ of $0$ contains not a single $x_n$ from the sequence. So $(x_n)$ does not converge at all in the box topology (if it would, $0$ from the weaker product topology is its only candidate limit and we saw that it fails.)
Your argument for the product topology and net convergence is almost correct: If $(x_\alpha)_{\alpha \in \Lambda}$ is a net that converges to $x$ in the product $X=\prod_{i \in I} X_i$, then $(x_i)_\alpha$ converges to $x_i$, for every $i$ simply because each projection $\pi_i$ is continuous and thus preserves net convergence (but the proof you give of it amounts to the same, but this is shorter). So this also holds in the box topology, but if we just assume that $(x_i)_\alpha \to x_i$ for all $i$, we can conclude $x_\alpha \to x$ in the product topology: Let $U=\prod_i U_i$ be a basic open product set, so $F=\{i \in I: U_i \neq X_i\}$ is finite and all $U_i$ are open in $X_i$. As $x_i \in U_i$ for each $i \in F$ we find $\alpha_i \in \Lambda$ such that
$$\forall \alpha \ge \alpha_i: (x_i)_\alpha \in U_i$$
using the convergence. Now there is no $\max(\alpha_i: i \in F)$, because in a directed set we know (and this is enough)
$$\exists \alpha_0: \forall i \in F: \alpha_i \le \alpha_0$$
as $F$ is finite (!). (We can do it for two (def. of directed set) and then induction gives a proof that we can do it for a finite subset  in a standard way, if one wants to be formal about it.) But writing $\max$ is in error, we "just" have a common upperbound.
But then for all $\alpha \ge \alpha_0$ we know that $\alpha \ge \alpha_i$ for all $i \in F$ and so $x_\alpha \in U$ for those $\alpha$ and we've shown convergence in the correct way. 
In the box topology we can have $F$ equal to $I$ and in infinite produts we thus need to find a common upperbound for infinitely many $\alpha_i$ and this is not always possible and the proof fails. The above counterexample shows that this is not an accident or lack of creativity, but an essential problem.
A: I think I got it. Actually, with a sucesion. Using $X_i=\mathbb{R}$ for every $i \in I$, I define $\{x_n\}_{n \in \mathbb{N}}\subset X$ as follows:
$$   
(x_n)_i = 
     \begin{cases}
       1/n \text{,  if  } n \leq i\\
       1 \text{, otherwise}\\
     \end{cases}
$$
So, clearly the projections of this net goes to $0_{X_i}$, but the original net never is close to $0_X$
