Product topology on uncountably index sets In the exercise I'm dealing with, the question says

Let $A$ be an uncountable set, for each $\alpha \in A$ let $X_\alpha = \{0,1\}$ with discrete topology. Let $X = \times_{\alpha \in A} X_\alpha$. Let $K = \{q \in X \mid q_\alpha = 0 \text{ except for a countable number of }\alpha\}$.
Show that $\overline{K} = X$.

I don't see it. $\overline{K}$ is the set of all limits of nets in $K$ that converge in $X$. For product sapce this means the net converges coordinate-wise. So how does any net in $K$ converge to $(\dots,1,\dots)$?
The exercise is from Bredon's Topology And Geometry, excecise 1.8.8. Actually the next part asks us to give an explict description of such net.
 A: So $K=\{q \in X: \text{supp}(q):=|\{\alpha: q_\alpha \neq 0\}| \le \aleph_0\}$
where as usual $|A| \le \aleph_0$ means that $A$ is at most countable.
If $U$ is a basic open subset of $X$, this means it is of the form (where the $\pi_\alpha$ are the continuous projections from $X$ onto $X_\alpha=\{0,1\}$:
$$U=\pi_{\alpha_1}^{-1}[\{p_1\}] \cap \pi_{\alpha_2}^{-1}[\{p_2\}] \cap \ldots \cap \pi_{\alpha_N}^{-1}[\{p_N\}] \text{ for finitely many } \alpha_1, \ldots \alpha_N \in J, \text{ and } p_1, \ldots p_N \in \{0,1\}$$
or equivalently 
$$U=\prod_{\alpha \in J} O_\alpha \text{ with finitely many } \alpha_1, \ldots \alpha_N \in J \text{ such that } \forall \alpha \notin \{\alpha_1, \ldots \alpha_N\}: O_\alpha = X_\alpha \text{ and } O_{\alpha_i} = \{p_i\}$$
Then the point $x \in X$ that satisfies $x_\alpha = 0$ for all $\alpha \notin \{\alpha_1, \ldots \alpha_N\}$ and $x_{\alpha_i}=p_i$ for $i=1,\ldots,N$ is clearly in $K \cap U$, so that $K$ intersects all basic open subsets of $X$, so $\overline{K}=X$. 
Now note that if $(x_n)$ is a sequence with all $x_n \in K$ and $x_n \to x$ in $X$, then $x \in K$ too: 
Let $A_n = \text{sup}(A_n)$ which is at most countable by definition.
Then $A = \bigcup_n A_n$ is also at most countable, and if $\alpha \notin A$ we know $\alpha \notin \text{supp}(x_n)$, so $(x_n)_\alpha=0$ for all $n$, so that $x_\alpha=0$ (as we have pointwise convergence in all coordinates) and so $\text{supp}(x) \subseteq A$ and $x \in K$.
So no sequence in $K$ can converge to $\mathbf{1}$, the constant 1 function in $X$.
But there is a net from $K$ that does converge: 
As index set use $[J]^\omega$, all subsets of $J$ that are at most countable, under inclusion as order relation, which is a directed set.
For $A \in [J]^\omega$ we define $y_A$ as the point of $X$ that is $1$ for $\alpha \in A$ and $0$ otherwise. 
For $U$ basic open as described above and containing $1$ (so all $p_i$ are $1$) we note that $A_0:=\{\alpha_1, \ldots, \alpha_N\} \in [J]^\omega$ and all $A \ge A_0$ implies that $y_A \in U$, so the definition of net convergence shows $y \to \mathbf{1}$
A: Hint. You need to prove that $K$ is dense in $X$, that is, meets every nonempty open set of $X$. What are the open sets of the product topology?
